ON A PROBLEM OF PARTITION OF NUMBERS

by
TIBERIU POPOVICIU
(Cluj)

1.

Let's consider equation
(1)

a_{1}x_{1}+a_{2}x_{2}+\ldots+a_{m}x_{m}=n

where $a_{1},a_{2},\ldots,a$ are $n$ given natural numbers. Assuming that $n$ is a non-negative integer, we will denote it by $\mathrm{N}\left(n;a_{1},a_{2},\ldots,a_{m}\right)$ the number of solutions in non-negative integers $x_{1},x_{2},\ldots,x_{m}$ of equation (1). For $n$ negative integer we define the symbol $\mathrm{N}\left(n;a_{1},a_{2},\ldots,a_{m}\right)$ as being equal to $(-1)^{m-}$ times the number of solutions in negative integers $x_{1},x_{2},\ldots,x_{m}$ of equation (1). $\mathrm{N}\left(n;a_{1},a_{2},\ldots,a_{n}\right)$ depends, apart from $n$ , and the coefficients $a_{1},a_{2},\ldots,a_{m}$ It is clear that the function $\mathrm{N}\left(n;a_{1},a_{2},\ldots,a_{m}\right)$ is symmetrical in $a_{1},a_{2},\ldots,a_{m}$ for each value of $n$ .

If $n$ is a positive integer, the number of solutions in positive integers of equation (1) is equal, based on the above, to $\left.(-1)^{m-1}\mathrm{\penalty 10000\ N}\left(-n;a_{1},a_{2},\ldots,a_{m}\right).{}^{1}\right)$

When there is no ambiguity about the coefficients $a_{1},a_{2},\ldots,a_{m}$ ,

⁰

Successive summations being extended to combinations $i_{1}$ MANY $1,i_{1},i_{2}$ MANY $2,\ldots,i_{1},i_{2},\ldots,i_{m-1}$ MANY $m-1$ of the indices $1,2,\ldots,m$ , we deduce

		$\displaystyle\mathrm{N}\left(-n;a_{1},a_{2},\ldots,a_{m}\right)=\Sigma\mathrm{N}\left(n;a_{i_{1}}\right)-\Sigma\mathrm{N}\left(n;a_{i_{1}},a_{2}\right)+\ldots+$
		$\displaystyle\quad+(-1)^{m-2}\Sigma\mathrm{\penalty 10000\ N}\left(n;a_{i_{1}},a_{i_{2}},\ldots,a_{l_{m-1}}\right)+(-1)^{m-1}\mathrm{\penalty 10000\ N}\left(n;a_{1},a_{2},\ldots,a_{m}\right)$

$\pi$ being a natural number.
we will write it shorter with $\mathrm{N}_{n}(n)$ , or even with $\mathrm{N}(n)$ , the number

\mathrm{N}\left(n;a_{1},a_{2},\ldots,a_{m}\right)

Index notation $m$ will be used especially when both equality (1) and the coefficients occur simultaneously $a_{1},a_{2},\ldots,a^{m}$ as well as analogous equations with less than $m$ unknowns, having as coefficients the terms from the beginning of the series $a_{1},a_{2},\ldots$ E.g. clues $m$ and $m-1$ distinguish the respective cases when we simultaneously consider equation (1) and equation $a_{1}x_{1}+a_{2}x_{2}+\ldots+a_{n-1}a_{n-1}=n$ We will use similar abbreviations for other functions as well. $n$ which also depend on the coefficients of equation (1). When necessary, we will always specify, for greater clarity, to which of equations (1) the abbreviated notations used refer.
2. - Based on Euler's research [1], the study of the number N(n) can be done using the generating functions

	$\displaystyle\mathrm{F}(\zeta)=\frac{1}{\left(1-\zeta^{a_{1}}\right)\left(1-\zeta^{a_{2}}\right)\cdots\left(1-\zeta^{a_{m}}\right)}=\sum_{n=0}^{\infty}\mathrm{N}(n)\zeta^{n}$		(2)
	$\displaystyle-\mathrm{F}(\zeta)=\sum_{n=1}^{\infty}\mathrm{N}(-n)\frac{1}{\zeta^{n}}$		(3)

which are obtained by developing the rational function $F(\zeta)$ once according to the increasing powers and once according to the decreasing powers of $\zeta$ .

Euler's method, based on formula (2), has been further developed by various researchers, especially by JJ Sylvester [2a]. This method uses the properties of the rational function $F(\zeta)$ and, in particular, its decomposition into simple fractions, as well as various properties of roots of different orders of unity. However, the number N(n) can also be studied through elementary considerations of number theory, a method that works well, at least in not too complicated cases. This method was applied in particular by K. Wei hrauch $[3a,3b]$ and, especially as regards the problems we are dealing with here, by Th. Skolem [4]. Both methods have their own advantages.
3. – In the present paper we propose to examine the following problem:

PROBLEM I. Given equation (1), find a polynomial $\mathrm{P}(n)$ of $n$ so that $\mathrm{N}(n)$ to be equal to the nearest whole number $\mathrm{P}(n)$ , whatever it is $n$ .

This problem has been studied in particular cases

		$\displaystyle m=3,a_{1}=1,a_{2}=2,a_{3}=3$
		$\displaystyle m=3,a_{1},=1,a_{2}=3,a_{3}=5$

by Sylvester [2a]. Th. Skolem [4] has also studied these cases

		$\displaystyle m=3,a_{1}=1,a_{2}=2,a_{3}=5$
		$\displaystyle m=4,a_{1}=1,a_{2}=2,a_{3}=3,a_{4}=5$

In the following we will completely solve the problem for $m=2$ and $m=3$ The case $m=1$ It is trivial and will be recalled in passing. We will also consider some more general cases.

Problem I (as well as the following problems II and II') allows, through certain simple particular solutions of it, the statement of some interesting properties on the number of solutions in non-negative integers or on the number of solutions in positive integers of equation (1).

Problem I is not always possible. In other words, given $m$ and the coefficients $a_{1},a_{2},\ldots,a_{n}$ , so equation (1), there is not always a polynomial $\mathrm{P}(n)$ so that the condition required by the problem is met. For a polynomial $\mathrm{P}(n)$ to verify the required condition it is necessary and sufficient to have

\begin{gathered}-\frac{1}{2}<\mathrm{P}(n)-\mathrm{N}(n)<\frac{1}{2}\\ |\mathrm{P}(n)-\mathrm{N}(n)|<\frac{1}{2}\end{gathered}

whatever $n$ If
in this case we know the value of $\mathrm{P}(n)$ , we can always unambiguously deduce the value of $\mathrm{N}(n)$ .
4. - Two other problems are closely related to problem I. One of these problems is the following:

PROBLEM II. - Given the equality (1), determine a polynomial $\mathrm{Q}(n)$ of $n$ so that $\mathrm{N}(n)$ to be equal to the whole contained in $\mathrm{Q}(n)$ whatever it is.

In other words, to determine a polynomial $Q(n)$ so that we have

\mathrm{N}(n)=[Q(n)]

(6)

whatever $n$ .
Equality (6) is equivalent to the inequalities

0\leqq\mathrm{Q}(n)-\mathrm{N}(n)<1

(7)

whatever $n$ In formula ( 6
) we have usually noted with [ $\alpha$ ] the entire contents of $\alpha$ , or the whole part of it $\alpha$ , or the largest integer $\leqq\alpha$ . Instead of [ $\alpha$ ] we can consider the whole that encompasses $\alpha$ , or the smallest integer $\geqq\alpha$ This number is equal to $-[-\alpha]$ We can then ask ourselves the question:

PROBLEM II'. - Given equation (1), determine a polynomial $\mathrm{S}(n)$ of $n$ just like that $\mathrm{N}(n)$ to be equal to the whole that includes $\mathrm{S}(n)$ , whoever it may be $n$ .

In other words, to determine a polynomial $S(n)$ so that we have (8)

\mathrm{N}(n)=-[-\mathrm{S}(n)]

whatever $n$ .

Equality (8) is equivalent to inequalities
(9)

0\leqq\mathrm{\penalty 10000\ N}(n)-\mathrm{S}(n)<1,

whatever $n$ If
problem I has a solution $\mathrm{P}(n)$ and problems II and II' have a solution because it is enough to take then $Q(n)=\mathrm{P}(n)+\frac{1}{2},\mathrm{\penalty 10000\ S}(n)=\mathrm{P}(n)-\frac{1}{2}$ so that inequalities (7) and (9) can be verified, since this follows simply from (4). We will see below that if one of the problems I, II, II' has a solution, the other two also have a solution, as well as what are the connections between the solutions of these problems.
5. - Finally, let us consider the more general problem:

PROBLEM III. - Given equation (1), determine a polynomial $\mathrm{R}(n)$ of $n$ so that the difference $\mathrm{R}(n)-\mathbb{N}(n)$ to be uniformly bordered.

In other words, determine a polynomial R(n) such that we have (10)

|\mathrm{R}(n)-\mathrm{N}(n)|<\mathrm{K}

whatever $n$ , K being an independent number of $n$
From inequalities (5), (7), (9) it is then seen that
Condition (10) is necessary for problems I, II, II' to have - solution .

We will see below how the solutions to problems I, II, II', when they exist, can be deduced from the polynomial $\mathrm{R}(n)$ .
6. - In the following we use some elementary properties of polynomials. Some of these properties result implicitly from the important property that a given polynomial is completely determined by a finite number of its values. No difficulty or confusion arises from the fact that the variable only goes through integer values. We also use the following:

LEMMA 1. - If the difference of two polynomials in $n$ is uniformly bounded, these polynomials differ by a constant.

We can accept this property here without proof.
From Lemma 1 it follows that any two solutions of one of the problems I, II, II' or III always differ by a constant. It can also be easily seen that the half-sum of two solutions is always a solution. So if $\mathrm{P}(n)$ is a solution to one of problems I, II, II' or III, all solutions of this problem are contained in the formula $\mathrm{P}(n)+\lambda$ , where $\lambda$ , belongs to a certain interval. In the case of the problem, it is a bracket belonging to an axis III, the range of its variation $\lambda$ is, obviously, the entire real axis. These properties will be specified below.
7. - - An observation naturally arises. Namely whether problems I, II, II', III become more general if we constrain $n$ to take only nonnegative values or only negative values. The answer is negative
, as follows from the following considerations. It is also clear that Lemma 1 also holds if we limit ourselves to only nonnegative values or only negative values of $n$ .

§ 1.

8.

– In this § we will deal with problem III. From the known results on equation (1) it can be deduced that for problem III to have a solution it is necessary and sufficient that the numbers $a_{1},a_{2},\ldots,a_{m}$ to be two primes between them. We will still give the proof of this property because, on the one hand, the necessity of the condition must be clearly shown, on the other hand, the proof of sufficiency leads to a direct procedure for calculating the polynomial $\mathrm{R}(n)$ , a procedure that should be specified.

We will first deal with the necessity of the condition. Here we will rely on the decomposition of the rational function (2) into simple fractions. In the cases $m=2,3$ we will also give direct demonstrations below.

We will assume $m>1$ , except unless otherwise specified.
9. - All roots of the denominator of the function $F(\zeta)$ are roots of unity and their order of multiplicity is, for each, at most $m$ In particular, the root 1 is of order $m$ of multiplicity. We then have the decomposition into simple fractions

F(\zeta)=\sum_{i=1}^{m}\sum_{(i)}\frac{\Gamma(\varepsilon,i)}{\left(1-\varepsilon^{i}\zeta,i\right.}

(11)

where the summation $\sum_{0}$ refers to all the different roots $\varepsilon$ , at least of the order of $i$ of multiplicity, of the denominator of the fraction $\mathrm{F}(\zeta)$ The constants Y ( $\varepsilon;$ i) are all different from zero, as follows from the fact that fraction (2) is irreducible and from the fact that the decomposition into simple fractions of a rational function is unique.

From (2) and (11) it follows that

\mathrm{N}(n)=\sum_{i=1}^{m}\mathrm{C}_{i}(n)\binom{n+i-1}{i-1}

(12)

where

C_{l}(n)=\sum_{(i)}\Gamma(\varepsilon;i)\varepsilon^{n},\quad i=1,2,\ldots,m

(13)

for anything $n$ nonnegative.
For brevity we will put

\delta=a_{1}a_{2}\ldots a_{m}

(14)

We then have $\varepsilon^{n^{\prime}}=\varepsilon^{n^{\prime\prime}}$ if $n^{\prime}\equiv n^{\prime\prime}(\bmod\delta)$ for all roots $\varepsilon$ From (13) we then deduce $\mathrm{C}_{i}\left(n^{\prime}\right)=\mathrm{C}_{i}\left(n^{\prime\prime}\right)$ for $n^{\prime}\equiv n^{\prime\prime}(\bmod\delta)$ , $i=1,2,\ldots,m$ ,

Doubly infinite strings
$\left\{C_{i}(n)\right\}$

\begin{gathered}\ldots,C_{i}(-1),C_{i}(0),C_{i}(1),\ldots\\ i=1,2,\ldots,m\end{gathered}

are therefore periodic with common period $\delta{}^{2}$ )
In particular the numbers $C(n),n=0,\pm 1,\pm 2,\ldots,i=1,2,\ldots,m$ take only a finite number of distinct values.
10. - We will now prove

LEMMA 2. - For problem III to have a solution it is necessary that for each $i=2,3,\ldots$ , m the numbers $\mathrm{C}_{1}(n)$ to take a single value.

Suppose that problem III has a solution but that the condition of the lemma is not satisfied. Then let $r$ his greatest value $i$ for which numbers $\mathrm{C}_{i}(n)$ take at least two different values. We then have $2\leqq r\leqq m$ . Either

C_{r}(x)\neq C_{r}(\xi).

(15)

If $\mathrm{R}(n)$ is a solution to problem III, we have in particular

|\mathrm{R}(\alpha+ns)-\mathrm{N}(\alpha+ns)|<\mathrm{K}

for anything $n$ whole.
But

\mathrm{N}(\alpha+ns)=\sum_{i=1}^{m}\mathrm{C}_{i}(\alpha)\binom{\alpha+ns+i-1}{i-1}

is a polynomial in $n$ and then, based on Lemma 1, we have

\mathrm{R}(\alpha+ns)=\mathrm{N}(\alpha+ns)+\lambda

whatever $n,\lambda$ being $o$ constant.
It then follows that

\mathrm{R}(n)=\sum_{i=1}^{m}\mathrm{C}_{i}(z)\binom{n+i-1}{i-1}+\lambda

(16)

Analogously we deduce
$\lambda^{\prime}$ being a constant.

\mathrm{R}(n)=\sum_{i=1}^{m}\mathrm{C}_{i}(\beta)\binom{n+i-1}{i-1}+\lambda^{\prime}

(17)

From (16), (17) we deduce

\sum_{i=1}^{m}\left[C_{i}(\alpha)-C_{l}(\beta)\right]\binom{n+i-1}{i-1}=\sum_{i=1}^{r}\left[C_{i}(\alpha)-C_{i}(\beta)\right]\binom{n+i-1}{i-1}=\lambda^{\prime}-\lambda

for anything $n$ .
² ) Instead of $\delta$ can you take the cmmmc of the numbers $a_{2},a_{2},\ldots,a_{m}$ .

However, this is impossible because, based on (15), the first term is a polynomial in $n$ of effective degree $r-1\geqq 1$ .
11. - We will now demonstrate

LEMMA 3. - For problem III to have a solution, it is necessary that the denominator of the fraction (2) does not have any multiple root different from 1.

Let us assume the opposite. Then there are at least two roots that are at least double and we have

C_{2}(n)=\Gamma(i;2)+\sum_{\begin{subarray}{c}(i)\\ \varepsilon\neq 1\end{subarray}}\Gamma(\varepsilon;2)\varepsilon^{n}

for anything $n$ , because one of the roots of e is equal to 1.
But, if $\varepsilon\neq 1$ , we have $\sum_{n=0}^{\delta-1}\varepsilon^{n}=0$ , so

\sum_{n=J}^{\delta-1}C_{2}(n)=\delta\Gamma(1;2)

However, based on Lemma 2, the numbers $G_{2}(n)$ are all equal and then it results $\mathrm{C}_{2}(n)=(1;2)$ what would it be $n$ We deduce from this that

\sum_{\begin{subarray}{c}(j)\\ \varepsilon\neq 1\end{subarray}}\mathrm{\penalty 10000\ T}(\varepsilon;\swarrow)\varepsilon^{\prime}=0,\quad n=0,1,\ldots

which implies, based on a well-known property of systems of linear equations, that $\Gamma(\varepsilon;2)=0$ for all roots $\varepsilon\neq 1$ . This is, however, impossible, based on an observation from No. 9. With this Lemma 3 is proven.
12. - The necessity of the sought condition is now proven because the fact that the denominator of the fraction $F(\zeta)$ has no multiple root different from 1 is equivalent to the fact that the coefficients $a_{1},a_{2},\ldots,a_{m}$ There are two paths, two primes between them.

It should be noted that the sufficiency of the condition follows from the preceding. If the numbers $a_{1},a_{2},\ldots,a_{m}$ there are two paths two primes between them formula (12) reduces to
(18)
and can be taken

\mathrm{N}(n)=\sum_{i=2}^{m}\mathrm{\penalty 10000\ T}(1;n)\binom{n+i-1}{i-1}+\mathrm{C}_{1}(n)

So finally we have

\mathrm{R}(n)=\sum_{i=2}^{m}\mathrm{\penalty 10000\ T}(1;i)\binom{n+i-1}{i-1}

THEOREM 1. - The necessary and sufficient condition for problem III to have a solution is that the numbers $a_{1},a_{2},\ldots,a_{m}$ to be two primes between them.

A reasoning analogous to that of No. 11 made on him $\mathrm{C}_{2}(n)$ , applied here to $\mathrm{C}_{1}(n)$ , shows us that the numbers $\mathrm{C}_{1}(n),n=0,\pm 1,\pm 2,\ldots$ cannot all be equal to each other unless 1 is the only root of the denominator of the fraction (2). This is equivalent to the fact that the numbers $a_{1},a_{2},\ldots,a^{m}$ are all equal to 1.

In the following we will give a direct proof of the sufficiency of the condition.

§ 2.

13.
- —
  
  Since we will be using variable functions $n$ relative to equations (1) with $m$ and $m-1$ unknowns, we will use the abbreviated notations in the sense explained in No. 1. Thus,

\mathrm{N}_{m}(n)=\mathrm{N}\left(n;a_{1},a_{2},\ldots,a_{m}\right),\mathrm{N}_{m-1}(n)=\mathrm{N}\left(n;a_{1},a_{2},\ldots,a_{m-1}\right)

as well as other analogous notations that are self-explanatory.
We will first prove
LEMMA 4. - We have the formula

\mathrm{N}_{m}\left(n+ka_{m}\right)-\mathrm{N}_{m}(n)=\sum_{i=1}^{k}\mathrm{\penalty 10000\ N}_{m-1}\left(n+ia_{m}\right)

(19)

whatever the natural number is $k$ , the integer n and the natural number $m>1$ .

The proof is done by counting the solutions of equation (1) in a well-determined way. We will distinguish three cases:
$1^{\circ}.0\geqq n$ . Then the non-negative integer solutions of the equation

a_{1}x_{1}+a_{2}x_{2}+\ldots+a_{m}x_{m}=n+ka_{m}

(20)

are of two kinds. Some are obtained from the non-negative integer solutions of the equation

a_{1}x_{1}+a_{2}x_{2}+\ldots+a_{m}x_{m}=n,

(21)

collecting $k$ the $x_{m}$ The others are solutions in which $x_{m}<k$ Formula (19) is obtained by giving $x_{m}$ in (20) successively the values $0,1,2,\ldots,k-1$ .
$2^{\circ}$ . - $ka_{m}\leqq n<0$ In this case there is a natural number $k^{\prime}\leqq k$ so that

-k^{\prime}a_{m}\leqq n<-\left(k^{\prime}-1\right)a_{m}.

The solutions in non-negative integers of equation (20) are obtained by giving $x_{m}$ successive values $0,1,\ldots,k-k^{\prime}$ and it turns out that

\mathrm{N}_{m}\left(n+ka_{m}\right)=\sum_{i=k^{\prime}}^{k}\mathrm{\penalty 10000\ N}_{m-1}\left(n+ia_{m}\right).

(22)

Also, any solution in negative integers of equation (21) is obtained by giving $x_{m}$ successive values $-1,-2,\ldots,-k^{\prime}+1$ and it turns out that

(-1)^{m-1}\mathrm{\penalty 10000\ N}_{m}(n)=(-1)^{m-2}\sum_{i=1}^{k^{\prime}-1}\mathrm{\penalty 10000\ N}_{m-1}\left(n+ia_{m}\right)

-\mathrm{N}_{m}(n)=\sum_{t=1}^{k^{\prime}-1}\mathrm{\penalty 10000\ N}_{m-1}\left(n+ia_{m}\right)

(23)

Adding equalities (22), (23) member by member, we deduce formula (19).

case $k^{\prime}=1\mathrm{nu}$ is excluded because then we obviously have $N_{m}(n)=0$ , $(m>1)$ .
$3^{\circ}.n<-ka_{m}$ In this case any solution in negative integers of equation (21) or is obtained from the solutions in negative integers of equation (20) by subtracting $k$ FROM $x_{m}$ or is there a solution in which $x_{m}\geqq-k$ . Giving in (21) to $x_{m}$ successive values $-1,-2,\ldots,-k$ , we deduce

(-1)^{m-1}\mathrm{\penalty 10000\ N}_{m}(n)-(-1)^{m-1}\mathrm{\penalty 10000\ N}_{m}\left(n+ka_{m}\right)=(-1)^{m-2}\sum_{i=1}^{k}\mathrm{\penalty 10000\ N}_{m-1}\left(n+ia_{m}\right)

which also returns to formula (19).
Lemma 4 is completely proven.
Important particular cases of formula (19) are
(24)

\mathrm{N}_{m}\left(n+a_{m1}\right)-\mathrm{N}_{m}(n)=\mathrm{N}_{m-1}\left(n+a_{m}\right)

which is obtained by taking $k=1$ and

\mathrm{N}_{m}(n+\delta)-\mathrm{N}_{m}(n)=\sum_{i=1}^{\delta^{\prime}}\mathrm{N}_{m-1}\left(n+ia_{m}\right)

(25)

where $\delta$ is defined by (14) and

\delta^{\prime}=a_{1}a_{2}\ldots a_{m-1},\quad\delta^{\prime}a_{m}=\delta

(26)

In what follows we will continue to use the abbreviated notations (14) and (26).
14. - We can now prove

LEMMA 5. - If the numbers $a_{1},a_{2},\ldots,a_{m}$ there are two primes between them we have, for any $n$ whole

\mathrm{N}_{n}(n)=\mathrm{R}_{n}(n)+\mathrm{G}_{n}(n),

(27)

where $\mathrm{R}(n)=\mathrm{R}\left(n;a_{1},a_{2},\ldots,a_{m}\right)$ is a polynomial of degree $m-1$ in $n$ and $\mathrm{G}_{m}(n)=\mathrm{G}\left(n;a_{1},a_{2},\ldots,a_{m}\right)$ forms an infinite double string $\left\{\mathrm{G}_{m}(n)\right\}$ periodically from time to time $\delta$ .

The proof of the lemma is done by complete induction, based on formula (25).

We first observe that, under the conditions of the lemma, we have

\Sigma^{*}\mathrm{G}_{m}(n)=\sum_{n=j}^{\delta-1}\mathrm{G}_{m}(n)=\mathrm{H}_{m}\left(a_{1},a_{2},\ldots,a_{m}\right),

(28)

where $\Sigma^{*}$ extends to a system of $\delta$ his values $n$ forming a complete system of debris (mode $\delta$ ). The sum (28) is independent of $n$ .

Let us now assume that the property is true in the case of equations (1) with $m-1$ unknowns and let us show that it will also be true in the case of equations (1) with m unknowns. We then have

\mathrm{N}_{m-1}(n)=\mathrm{R}_{n-1}(n)+\mathrm{G}_{n-1}(n)

(29)

the two terms in the second member satisfying the conditions of the lemma.

\mathrm{R}_{m}(n+\delta)-\mathrm{R}_{m}(n)=\sum_{i=1}^{\delta^{\prime}}\mathrm{R}_{n-1}\left(n+ia_{m}\right)+\mathrm{H}_{m-1}\left(a_{1},a_{2},\ldots,a_{m-1}\right)

2.

Taking into account (25) and (20) we deduce

\mathrm{N}_{m}(n+\delta)-\mathrm{N}_{m}(n)=\mathrm{R}_{m}(n+\delta)-\mathrm{R}_{m}(n)

Putting the slides

\mathrm{G}_{m}(n)=\mathrm{N}_{m}(n)-\mathrm{R}_{m}(n)

(31)

the property is demonstrated.
Finally, for $m=1$ the lemma is true. To see this it is enough to take $\mathrm{R}\left(n;a_{1}\right)=0$ and we will then have $\mathrm{G}\left(n;a_{1}\right)=1$ or 0 as appropriate $a_{1}$ divide by $n$ or does not divide by $n$ .

During the demonstration that the numbers $a_{1},a_{2},\ldots,a_{m}$ are two primes between them, it occurs by applying formula (28). Namely the sum $\Sigma^{*}\mathrm{G}_{m-1}(n)$ extends here to the values $n+ia_{n},i=1,2,\ldots,\delta^{\prime}$ his/hers $n$ , values that form a complete system of solutions (mode $\delta^{\prime}$ ) because the numbers $a_{m},a_{1}a_{2}\ldots a_{m-1}$ are prime to each other. So the fact that the numbers $a_{1},a_{2},\ldots,a_{n}$ are two by two primes between them is equivalent to the fact that the numbers $a_{i},a_{1}a_{2}\ldots a_{i-1}$ are prime among themselves, whatever they are $i=2,3,\ldots,m$ .

With this the sufficiency of the condition of Theorem 1 is demonstrated.
15. - Let us suppose then that the numbers $a_{1},a_{2},\ldots,a_{m}$ are two primes among themselves. In formula (27) the polynomial $\mathrm{R}_{m}^{2}(n)$ is determined only by some additive constant. Our considerations in the following §§ are not influenced by this indeterminacy of the terms $\mathrm{R}_{m}(n)$ and $\mathrm{G}_{m}(n)$ , but the fact that we have this indeterminacy simplifies the exposition in certain cases.

To appreciate its structure $\mathrm{N}_{m}(n)$ and especially to calculate $\mathrm{R}_{m}(n)$ it is useful to conveniently fix the additive constant in this term. For this we will "normalize" the polynomial R ( $n$ ) through the condyle
(32) $\quad\mathrm{R}_{m}(0)=\mathrm{R}\left(0;a_{1},a_{2},\ldots,a_{m}\right)=0,\quad(m\geqq 1)$ .

By this condition the polynomial $\mathrm{R}_{n}(n)$ , so the periodic term $\mathrm{G}_{m}(n)$ from formula (27) are completely determined. If $m>1$ , the polynomial
$\mathrm{G}_{m}(n)$ is of the form ³ )

\mathrm{R}_{m}(n)=\frac{1}{\delta}\sum_{j=1}^{m-1}\mathrm{\penalty 10000\ A}_{j}^{(n)}n^{\prime}

where

\left.\mathrm{A}_{j}^{(m)}=\mathrm{A}_{j}^{(m)},a_{1},a_{2},\ldots,a_{n}\right),j=1,2,\ldots,m-1

are independent coefficients of $n$
We will now prove LEMMA
6. - If the numbers $a_{1},a_{2},\ldots,a_{n}$ are two primes between them, the coefficients of the polynomial $\mathrm{R}_{n}(n)$ , assumed to be normalized by condition (32), as well as $\mathrm{C}_{m}(n)$ are symmetric functions of $a_{1},a_{2},\ldots,a_{m}$ for any value of $n$ .

In what respects are the coefficients of the polynomial $\mathrm{R}_{m}(n)$ it is obvious enough to demonstrate that $\mathrm{A}_{j}^{(m)},j=1,2,\ldots,m-1$ are symmetric. The proof of the lemma is based on the observation that $\mathrm{N}_{n}(n)$ is symmetrical with respect to $a_{1},a_{2},\ldots,a_{n}$ .

From (27) we deduce, for a $n$ that,

\delta\mathrm{N}_{m}(n+i\delta)=\sum_{j=1}^{m-1}\mathrm{\penalty 10000\ A}_{j}^{(n)}(n+i\delta)^{\gamma}+\delta\mathrm{G}_{m}(n),\quad i=0,1\ldots,m-1

(33)

These relationships form a system of $m$ linear equations with $m$ unknown $\mathrm{A}_{1}^{(m)}A_{2}^{(m)},\ldots,\mathrm{A}_{m-1}^{(m)},\mathrm{G}_{m}(n)$ All coefficients of this system are symmetric functions of $a_{1},a_{2},\ldots,a_{m}$ and Lemma 6 results from solving the system. It is also easy to see that the determinant of the system is nonzero.

Referring to formula (28) we see that, if we have (32), the function $\mathrm{H}_{n}\left(a_{1},a_{2},\ldots,a_{m}\right)$ is also a symmetric function of $a_{1},a_{2},\ldots,a_{n}$ This property results, as we will see below, also directly from the symmetry of the coefficients $\mathrm{A}_{j}^{(m)}$ .

System (33) allows us to calculate the coefficients of the polynomial $\mathrm{R}_{n}(n)$ according to the numbers $\mathrm{N}_{m}(n)$ . It is unnecessary to write the corresponding formulas because they will not be used here. We will calculate these coefficients using formula (30). However, first it is necessary to establish some formulas used below.

§ 3.

16.
- —
  
  Bernoulli numbers are defined by the recurrence relations [5]

\mathrm{B}_{0}=1,\sum_{v=j}^{s}\binom{s}{v}\mathrm{\penalty 10000\ B}_{v}=\mathrm{B}_{s}s=2,3,\ldots

(34)

3.

Highlighting the factor $\frac{1}{\delta}=\frac{1}{a_{1}a_{2}\ldots a_{m}}$ simplifies the exposition, as we will see below.

From here we deduce and

\mathrm{B}_{0}=1,\frac{\mathrm{\penalty 10000\ B}_{j}}{j!}=-\sum_{v=0}^{j-1}\frac{\mathrm{\penalty 10000\ B}_{v}}{v!(j+1-v)!},\quad j=1,2,\ldots

(35)

These formulas allow us to successively calculate the numbers $\mathrm{B}_{y}$ We thus obtain

B_{0}=1,B_{1}=-\frac{1}{2},B_{2}=\frac{1}{6},B_{4}=-\frac{1}{30},B_{6}=\frac{1}{42},B_{8}=-\frac{1}{30},

It is known that all Bernoulli numbers with odd indices and $>1$ are null, and those with even indices and $>0$ are alternately positive and negative. More precisely

\mathrm{B}_{4v-2}>0,\quad\mathrm{\penalty 10000\ B}_{4v}<0,\quad v=1,2,\ldots

Calculating the difference

\sum_{v=0}^{s-1}\frac{B_{v}}{v|(s-v)|}-s\sum_{v=0}^{s}\frac{B_{v}}{v|(s+1-v)|},s\geqq 2

and taking into account (35) we deduce

\frac{1}{(s+1)!}-\sum_{v=1}^{s}\frac{(v-1)B_{v}}{v!(s+1-v)!}=\frac{B_{s}}{s!},\quad s\geqq 2

(36)

We will also need to evaluate the amounts

\mathrm{I}_{s}=\sum_{v=1}^{s}\frac{\mathrm{\penalty 10000\ B}_{v}}{v\mathrm{l}}\cdot\frac{\mathrm{\penalty 10000\ B}_{s-v}}{(s-v)!},\quad s=0,1,\ldots

namely, they will demonstrate that

\mathrm{I}_{0}=1,\sum_{v=0}^{s}\frac{\mathrm{\penalty 10000\ B}_{v}}{v!}\cdot\frac{\mathrm{B}_{s-v}}{(s-v)!}=-\frac{\mathrm{B}_{s-1}}{(s-1)!}-\frac{(s-1)\mathrm{B}_{s}}{s!},s=1,2,\ldots

(37)

For this we observe that, if we take into account (35) and if we assume $s>0$ , we deduce

	$\displaystyle l_{s}=\sum_{v=0}^{s}\frac{\mathrm{\penalty 10000\ B}_{v}}{v!}\cdot\frac{\mathrm{B}_{s-v}}{(s-v)!}$	$\displaystyle=\frac{\mathrm{B}_{s}}{s!}-\sum_{v=0}^{s-1}\frac{\mathrm{\penalty 10000\ B}_{v}}{v!}\left[\sum_{\mu=0}^{s-v-1}\frac{\mathrm{\penalty 10000\ B}_{\mu}}{\mu!(s+1-v-\mu)!}\right]=$
		$\displaystyle=\frac{\mathrm{B}_{s}}{s!}-\sum_{j=1}^{s-1}\left[\sum_{v=0}^{j}\frac{\mathrm{\penalty 10000\ B}_{v}}{v!}\cdot\frac{\mathrm{B}_{j-v}}{(j-v)!(s-j+1)!}\right]$

from where

\mathrm{I}_{s}=\frac{\mathrm{B}_{s}}{s!}-\sum_{j=0}^{s-1}\frac{\mathrm{I}_{j}}{(s-j+1)!}\cdot s=1,2,\ldots

(38)

Formula (37) is proved by complete induction. It is enough to assume $s>1$ , because for $s=0,1$ the formulas are verified directly immediately. Assuming that the formula is true for $I_{0},I_{1},\ldots,I_{c-1}$ I must show that it will be true for $l_{s}$ This follows from formula (38), noting that

\sum_{j=1}^{s-1}\frac{\mathrm{l}_{j}}{(s-j+1)!}=\frac{1}{(s+1)!}-\sum_{j=1}^{s-1}\frac{1}{(s-j+1)!}\left[\frac{\mathrm{B}_{j-1}}{(\jmath-1)!}+\frac{(-j1)\mathrm{B}_{j}}{j!}\right]

and taking into account formulas (35), (36).
17. - Let us now consider the polynomials in $t$

\Phi_{s}=\Phi_{s}(t)=\sum_{v=0}^{s}\frac{\mathrm{\penalty 10000\ B}_{v}}{v!}t^{v},s=0,1,\ldots

We propose to calculate the coefficient of $t^{s}$ in the polynomial $\Phi_{s}^{s+1}$ , that is, the number of coefficients $\boldsymbol{\Phi}_{s}^{s+1}$ , agreeing to note with $\operatorname{coef}_{k}\Phi$ his coefficient $t^{k}$ in the polynomial $\Phi$ of $t$ .

For this, let us also consider the polynomials

\mathrm{E}_{s}=\mathrm{E}_{s}(1)=\sum_{v=0}^{s}\frac{t^{v}}{v!},\quad s=0,1,

If we observe it $B_{1}=-\frac{1}{2}$ and if we take into account formulas (34) we have

\Phi_{s}\mathrm{E}_{s}\equiv l+\Phi_{s}\quad\left(\bmod t^{s+1}\right)

(39)

Where from
?

\Phi_{\mathrm{s}}^{s+1}\equiv-t\Phi_{s}^{s}+\Phi_{s}^{s+1}\mathrm{E}_{s}\quad\left(\bmod t^{s+1}\right)

and we deduce

t\Phi_{s}^{s}\equiv t\Phi_{s-1}^{s}\quad\left(\bmod t^{s+1}\right)

(40)

\operatorname{coef}_{s}\Phi_{s}^{s+1}=-\operatorname{coef}_{s-1}\Phi_{s-1}^{s}+\operatorname{coef}_{s}\Phi_{s}^{s+1}E_{s}

Taking into account formulas (37) we deduce ( $\Phi_{s}^{\prime}$ is the derivative of $\Phi_{s}$ )

t\Phi_{s}^{\prime}\equiv\Phi_{s}\left(1-t-\Phi_{s}\right)\quad\left(\bmod t^{s+1}\right)

and if we also take into account (35),
whence

t\Phi_{s}^{\prime}-\Phi_{s}\equiv-\Phi_{s}^{2}E_{s}\quad\left(\bmod t^{s+1}\right)

i\Phi_{s}^{\prime}\Phi_{s}^{s-1}-\Phi_{s}^{s}\equiv-\Phi_{s}^{s+1}E_{s}\quad\left(\bmod\iota^{s+1}\right)

We therefore have

\operatorname{coef}_{s}\left[t\Phi_{s}^{\prime}\Phi_{s}^{s-1}-\Phi_{s}^{s}\right]=-\operatorname{coef}_{s}\Phi_{s}^{s+1}\mathrm{E}_{s}

But the first member is null because $s\Phi_{s}^{\prime}\Phi_{s}^{s-1}$ is the derivative of the polynomial $\Phi_{s}^{s}$ We therefore deduce from (40)

\operatorname{coef}_{s}\Phi_{s}^{s+1}=-\operatorname{coef}_{s\rightarrow 1}\Phi_{s-1}^{s}

Where, after all,

\operatorname{coef}_{s}\Phi_{s}^{s+1}=(-1)^{s},\quad s=0,1,\ldots

(41)

18.
- —
  
  Let's note with $\left[\gamma_{1},\gamma_{2},\ldots,\gamma_{j}\right]_{m}$ symmetric function $\Sigma a_{1}{}^{1}\cdot u_{2}^{\gamma_{2}}\ldots a_{j}^{\gamma_{j}}$ relative to variables $a_{1},a_{2},\ldots,a_{m}$ So we have

\left[\gamma_{1},\gamma_{2},\ldots,\gamma_{j}\right]_{m}=\sum a_{1}^{\gamma_{1}}a_{2}^{\gamma}\cdots\gamma_{j}

(42)

Here the exponents $\gamma_{1},\gamma_{2},\ldots,\gamma_{j}$ are natural numbers $(j\leqq m)$ and we can, for the sake of notation, always assume $\gamma_{1}\leqq\gamma_{2}\leqq\leqq\ldots\leqq\gamma_{j}$ .

In the case where there is no ambiguity over the variables whose symmetric functions are considered, we can write it more simply with $\left[\gamma_{1},\gamma_{2},\ldots,\gamma_{j}\right]$ function $\left[\gamma_{1},\gamma_{2},\ldots,\gamma_{j}\right]_{m}$ , suppressing the index $m$ .

Let us consider the symmetric function

	$\displaystyle\varphi_{s}^{(l+1)}\left(a_{1},c_{2},\ldots,a_{m}\right)=$		(43)
	$\displaystyle=\sum_{(s)}\left(\frac{B_{1}}{1!}\right)^{a_{1}}\left(\frac{B_{2}}{2!}\right)^{a_{2}}\ldots\left(\frac{B_{s}}{s!}\right)^{a_{s}}[1,\underbrace{1,\ldots,}_{\alpha_{1}}1,\underbrace{2,2,\ldots,}_{\alpha_{2}}2,\ldots,\underbrace{s,\ldots,s}_{\alpha_{s}}]_{m}$

where the summation $\sum_{(s)}$ refers to all solutions in non-negative integers $\alpha_{1},\alpha_{2},\ldots,\alpha_{s}$ of the equation $\alpha_{1}+2\alpha_{2}+\ldots+s\alpha_{s}=s$ We obviously also have $\alpha_{1}+\alpha_{2}+\ldots+\alpha_{s}\leqq m$ . In particular

\varphi_{0}^{(m)}\left(a_{1},a_{2},\ldots,a_{m}\right)=1

whatever $m$ .
Moreover, in our considerations only the functions (43) will be involved in which $s<m$ .

The function (43) is a symmetric and homogeneous polynomial of degree $s$ in $a_{1},a_{2},\ldots,a_{m}$ We have the immediate formula

\psi_{s}^{(m)}\left(a_{1},a_{2},\ldots,a_{m}\right)=\sum_{v=0}^{s}\frac{B_{v}}{vl}a_{m}^{v}\varphi_{s-v}^{(m-1)}\left(a_{1},a_{2},\ldots,a_{m-1}\right)

(44)

If there is no ambiguity about the variables whose symmetric functions are considered, we can write it more briefly with the function $\varphi_{s}^{(n)}\left(a_{1},a_{2},\ldots,a_{m}\right)$ We can then write

\varphi_{s}=\sum_{(s)}\left(\frac{B_{1}}{11}\right)^{\alpha_{1}}\left(\frac{B_{2}}{21}\right)^{\alpha_{2}}\ldots\left(\frac{B_{s}}{s1}\right)^{\alpha_{s}}[1,\underbrace{1,\ldots,1}_{\alpha_{1}}2\underbrace{2,2,\ldots,2}_{\alpha_{2}},\ldots,\underbrace{s,s,\ldots,s}_{\alpha_{s}}]

(45)

Because $\mathbf{B}_{v}$ , is null for $v$ odd and $>1$ , in $\varphi_{s}$ only the terms in which it actually appears are $\alpha_{i}=0$ for $i$ odd and $>1$ It follows from this that if $s$ is odd in $\varphi_{s}$ actually only appear in terms in which we have $\alpha_{1}>0$ .

From the structure of functions (45) a practical rule results, given by JJ Sylvester $[2b]$ , for the formation of functions $\varphi_{s}$ from previous functions $\varphi_{0},\varphi_{1},\ldots,\varphi_{s-1}$ For this from $\varphi_{l}(0\leqq i\leqq s-1)$ we only take the sum of the terms in which the brackets [] contain only numbers $\geq s-i$ We complete with a $s-i$ each of these brackets and multiply the result by $\frac{\mathrm{B}_{s-i}}{(s-i)!}$ We thus obtain a sum $\varphi_{i}^{*}$ . In particular $\varphi_{0}^{*}=\frac{\mathrm{B}s}{s!}[s]$ We then have

\varphi_{s}=\sum_{i=0}^{s-1}\varphi_{i}^{*}.

If now $s$ it is odd we have obviously $\varphi_{i}^{*}=0$ if $i$ este par şı $<s-1$ But we also have $\varphi_{i}^{*}=0$ if $i$ is odd and $<s-1$ because then in $\varphi_{i}$ there is no term with a bracket [] in which all numbers are $\geqq s-i$ If $s$ is odd so we have

\varphi_{s}=\varphi_{s-1}^{*}.

If $s$ it seems we have it obvious $\varphi_{i}^{*}=0$ for $i$ odd and $<s-1$ We also have $\varphi_{i}^{*}=0$ for $i$ hair and $<\frac{s}{2}$ because then there is no $\varphi_{i}$ no brackets [] where we only have numbers $\geqq s-i$ In his case $s$ so we have

\varphi_{s}=\varphi_{0}^{*}+\sum_{i=\left[\frac{s-1}{4}\right]+1}^{\frac{s}{2}-1}\varphi_{2i}^{*}+\varphi_{s-1}^{*}.

With the help of these rules we can successively construct the functions $\varphi_{s}$ We give below these functions for $s\leqq 9$ .

	$\displaystyle\varphi_{0}=1,\quad\varphi_{1}=-\frac{1}{2}[1],\quad\varphi_{2}=\frac{1}{12}([2]+3[1]),$
	$\displaystyle\varphi_{3}=-\frac{1}{24}([2]+3[11]),$
	$\displaystyle\varphi_{4}=\frac{1}{6!}(-[4]+5[2]+5[12]+5[111]),$
	$\displaystyle\varphi_{5}=-\frac{1}{2.6!}(-[4]+5[22]+5[112]+5[1111]),$
	$\displaystyle\varphi_{6}=\frac{1}{12.7!}(2[6]-7[4]-1[14]+5[22]+05[122,]+$
	$\displaystyle+15[1112]+45[11111]),$

		$\displaystyle\varphi_{7}=-\frac{1}{3.8!}(2[6]-7[24]-1[114]+5[222]+$
		$\displaystyle+05[1122]+15[11112]+45[111111])$
		$\displaystyle\varphi_{8}=\frac{1}{10!}(-3[8]+7[4]+0[6]-5[24]+0[16]+$
		$\displaystyle+75[222]-05[124]-15[1114]+$
		$\displaystyle+25[1222]+575[11122]+$
		$\displaystyle+725[111112]+4175[1111111])$
		$\displaystyle\varphi_{9}=-\frac{1}{2.10!}(-3[8]+7[44]+0[26]-5[224]+$
		$\displaystyle+0[116]+75[2222]-05[1124]-$
		$\displaystyle+15[11114]+25[11222]+$
		$\displaystyle+575[111122]+725[1111112]+$
		$\displaystyle+4175[11111111]).$

Using formula (42) all brackets [] can be expressed using symmetric functions of the variables on which the functions depend. $\varphi_{s}$ If we do $a_{1}=a_{2}=\ldots=a_{m}=1$ , we have
$[\underbrace{1,1,\ldots,}_{\alpha_{1}}1,\underbrace{2,\ldots,2}_{\alpha_{2}},\ldots,\underbrace{s,s,\ldots,s}_{\alpha_{s}}]_{m}=\frac{m!}{\alpha_{1}\left|\alpha_{2}\right|\ldots\alpha_{s}!\left(m_{1}-\alpha_{1}-\alpha_{2}-\ldots-\alpha_{s}\right)\mid}$ . and we deduce

\psi_{s}^{(m)}(1,1,\ldots,1)=

$=\sum_{(s)}\left(\frac{B_{1}}{1!}\right)^{\alpha_{1}}\left(\frac{B_{2}}{2!}\right)^{\alpha_{2}}\ldots\left(\frac{B_{s}}{s!}\right)^{\alpha_{s}}\frac{m!}{\alpha_{1}!\alpha_{2}!\ldots\alpha_{s}!\left(m-\alpha_{1}-\alpha_{2}-\ldots-\alpha_{s}\right)!}=\operatorname{coef}_{s}\Phi_{s}^{m}$
Taking into account (41) we have therefore
(46)

\varphi_{m-1}^{(m)}(1,1,\ldots,1)=(-1)^{m-1},m=1,2,\ldots

formula that will be used below.

\S 4.

19.
- —
  
  Let's return to calculating the coefficients of the polynomial $\mathrm{R}_{m}(n)$ .

By a simple identification, we can establish, based on formulas (34), the following identity ⁴ )
4). We have

x^{j+1-v}-\left(x-a_{ln}\right)^{j+1-v}=\sum_{\mu=0}^{j-v}(-1)^{j-v-\mu}\binom{j+1-v}{\mu}x^{\mu}a_{m}^{j+1-v-\mu}

and the second member of the formula is written

x^{j}=\sum_{v=0}^{j}(-1)^{v}\binom{j}{v}_{\mathrm{B}_{v}a_{m}^{v-1}}\frac{x^{j+1-v}-\left(x-a_{m}\right)^{i+1-v}}{j+1-v}

Giving him $x$ successive values $n+ia_{m},i=1,2,\ldots,\delta^{\prime}$ and adding term by term we deduce

\sum_{i=1}^{\delta^{\prime}}\left(n+ia_{m}\right)^{j}=\sum_{v=0}^{j}(-1)^{v}\binom{j}{v}\mathrm{\penalty 10000\ B}_{v}a_{m}^{v-1}\frac{(n+\delta)^{i+1-v}-n^{j+1-v}}{j+1-v}

where we use the abbreviation (26).

		$\displaystyle\sum_{i=1}^{\delta^{\prime}}\mathrm{R}_{m-1}\left(n+ia_{m}\right)=\frac{1}{\delta}\sum_{j=1}^{m-2}\mathrm{\penalty 10000\ A}_{i}^{(m-1)}\left[a_{m}\sum_{i=1}^{\delta^{\prime}}\left(n+ia_{m}\right)^{j}\right]=$
		$\displaystyle\left.=\frac{1}{\delta}\sum_{j=1}^{m-2}\mathrm{\penalty 10000\ A}_{j}^{(m-1)}\left\lvert\,\sum_{v=0}^{j}(-1)^{\nu}\binom{j}{v}\mathrm{\penalty 10000\ B}_{v}a_{m}^{v}\frac{(n+\delta)^{j+1-v-n}-n^{j+1-v}}{j+1-v}\right.\right]=\sum_{v=1}^{m-2}(-1)^{\nu}\mathrm{B}_{v}v_{m}^{v}\mathrm{\penalty 10000\ A}_{v}^{(m-1)}+$
		$\displaystyle\quad+\frac{1}{\delta}\sum_{j=2}^{m-1}\frac{1}{j!}\left[\sum_{v=0}^{m-1-j}(-1)^{\nu}\frac{\mathrm{B}_{v}}{\nu!}a_{m}^{v}(j+v-1)!\mathrm{A}_{j+v^{-1}-1}^{(m-1)}\right]\left[(n+\delta)^{j}-n^{j}\right].$

We then have

\mathrm{R}_{m}(n+\delta)-\mathrm{R}_{m}(n)=\frac{1}{\delta}\sum_{j=1}^{m-1}\mathrm{\penalty 10000\ A}_{j}^{(m)}\left[(n+\delta)^{i}-n^{j}\right]

Identifying in identity (27) the coefficients of the polynomials $(n-\dot{x})^{j}--n,j=1,2,\ldots,m-1$ , which forms an independent linear system, we have

\mathrm{A}^{(m)}=\sum_{v=1}^{m-2}(-1)^{v}\mathrm{\penalty 10000\ B}_{v}a_{m}^{v}\mathrm{\penalty 10000\ A}_{v}^{(m-1)}+\mathrm{H}_{m-1}\left(a_{1},a_{2},\ldots,a_{m-1}\right),

(47)

\mathrm{A}_{j}^{(m)}=\frac{1}{j!}\sum_{v=0}^{m-j}\frac{(-1)^{v}\mathrm{\penalty 10000\ B}_{v}}{v!}a_{m}^{v}(j+v-1)!\mathrm{A}_{j+v-1}^{(m-1)},\quad j=2,3,\ldots,m-1.

(48)

If $m=2$ formula (30) reduces to

\begin{gathered}\sum_{v=0}^{j}\frac{1}{j+1-v}(-1)^{v}\binom{j}{v}\mathrm{\penalty 10000\ B}_{v}a_{m}^{v-1}\left[\sum_{\mu=0}^{j-v}(-1)^{j-v-\mu}\binom{j+1-v}{\mu}x^{\mu}a_{m}^{j+1-v-\mu}\right]=\\ =\sum_{\mu=0}^{j}\frac{(-1)^{i-\mu}j!}{\mu!(j+1-\mu)!}x^{\mu}a_{m}^{j-\mu}\left[\sum_{v=0}^{j-\mu}\binom{j+1-\mu}{v}\mathrm{\penalty 10000\ B}_{v}\right]=x^{j}\end{gathered}

because, based on formula (34), the sum $\sum_{v=0}^{j-\mu}\binom{j+1-\mu}{v}\mathrm{\penalty 10000\ B}_{v}$ is equal to 1 or 0 if $\mu=j$ resp. $\mu<j$ .

\mathrm{R}_{2}\left(n+a_{1}a_{2}\right)-\mathrm{R}_{2}(n)=\mathrm{H}_{1}\left(a_{1}\right)

and formulas (47), (48) reduce to the single formula

\mathrm{A}_{1}^{(2)}=\mathrm{H}_{1}\left(a_{1}\right)

(49)

It can now be seen that the symmetry of $H_{m-1}\left(a_{1},a_{2},\ldots,a_{m-1}\right)$ results from that of the coefficients $\mathrm{A}_{j}^{(m)}$ Indeed, from (47) it follows

\mathrm{H}_{m-1}\left(a_{1},a_{2},\ldots,a_{m-1}\right)=\mathrm{A}_{1}^{(m)}\left(a_{1},a_{2},\ldots,a_{m-1},0\right)

(

(50

)

20.
- —
  
  The final result of the coefficient calculation $\mathrm{A}_{j}^{(n)}$ and the function $\mathrm{H}_{m-1}\left(a_{1},a_{2},\ldots,a_{m-1}\right)$ it is stated thus

THEOREM 2. - Functions $\mathrm{A}_{j}^{(m)},j=1,2,\ldots,m-1$ of $a_{1},a_{2},\ldots,a_{m}$ are symmetric and homogeneous polynomials of the respective degrees $m-1-j$ , $j=1,2,\ldots,m-1$ Also $H_{\eta_{-1}}\left(a_{1},a_{2},\ldots,a_{m-1}\right)$ is a symmetric and homogeneous polynomial of degree $m-2$ , We have

	$\displaystyle\mathbb{A}_{j}^{(m)}\left(a_{1},a_{2},\ldots,a_{m}\right)=\frac{(-1)^{m-1-1}}{9!}q_{m-1-1}^{(m)}\left(a_{1},a_{2},\ldots,a_{m}\right)$		(51)
	$\displaystyle j=1,2,\ldots,m-1$
	$\displaystyle\mathrm{H}_{m-1}\left(a_{1},a_{2},\ldots,a_{m-1}\right)=(-1)^{m-2}q_{m-2}^{(m-1)}\left(a_{1},a_{2},\ldots,a_{m-1}\right)$		(52)

The proof is done by complete induction based on the recurrence formulas (47), (48). For this we first observe that

\mathrm{H}_{m}(1,1,\ldots,1)=1,m=1,2,\ldots

(53)

Indeed, in this case, from (28) it follows
(54)

\mathrm{G}_{m}(0)=\mathrm{H}_{m}(1,1,\ldots,1)

But $\mathrm{N}_{m}(0)=1,\mathrm{R}_{m}(0)=0$ and from (27) it follows that $\mathrm{G}_{m}(0)=1$ Taking into account (54) we deduce formula (53).

From formula (49) we deduce that $H_{1}\left(a_{1}\right)$ is independent of $a_{1}$ , because it must be a symmetric function of $a_{1}$ and $a_{2}$ . $\mathrm{H}_{3}\left(a_{1}\right)$ therefore reduces to a constant which, based on formula (53), is equal to 1. We have

\mathrm{A}_{1}^{(2)}\left(a_{1},a_{2}\right)=1,\mathrm{H}_{1}\left(a_{1}\right)=1

This means that formulas (51), (52) are verified for $m=2$ .

Let us assume that the formulas are true for $\mathrm{A}_{j}^{(n-i)},j=1,2$ , $\ldots,m-2(m>2)$ and show that they will also be true for Afs $j=1,2,\ldots,m-1,\mathrm{H}_{n-1}\left(a_{1},a_{2},\ldots,a_{m-1}\right)$ .

By hypothesis we have

\mathrm{A}_{j}^{(n-1)}\left(a_{1},a_{2},\ldots,a_{m-1}\right)=\frac{(-1)^{m-2-j}}{j!}\psi_{m-2-j}^{(n-1)}\left(a_{1},a_{2},\ldots,a_{m-1}\right)

Taking into account formulas (44), (48) we immediately deduce that formulas (51) are verified for $j=2,3,\ldots,m-1$ It remains to establish formula (51) for $j=1$ . We will establish this formula together with formula (52).
21. – For this we will use the following

LEMMA 7. - If the function $\chi\left(x_{1},\mathrm{x}_{2},\ldots,x_{n-1}\right)$ , depending only on the first $m-1$ variable $x_{1},x_{2},\ldots x_{n-1}$ , is a symmetric function of all the $m$ variable $x_{1},x_{2},\ldots,x_{m}$ , then it reduces to a constant.

First it is clear that the function $\chi\left(x_{1},x_{2},\ldots,x_{n-}\right)$ is symmetrical with respect to the $m-1$ variable $x_{1},x_{2},\ldots,x_{m=1}$ We then deduce, from the other symmetry conditions, that

\chi\left(x_{1},x_{2},\ldots,x_{m-1}\right)=\chi\left(x_{1},x_{2},\ldots,x_{m-2},x_{m}\right)

and doing here $x_{m}=a$ ,

\chi\left(x_{1},x_{2},\ldots,x_{m-1}\right)=\chi\left(x_{1},x_{2},\ldots,x_{m-2},a\right),

(55)

$a$ being one of the values that variables can take $x_{1},x_{2},\ldots,x_{m}$ .
Formula (55) shows us that the function $\chi$ it only depends on the first $m-2$ variable $x_{1},x_{2},\ldots,x_{..,-2}$ . Repeating the reasoning, it is found that the function $\chi$ reduces to a conscious ⁵ ). We therefore have
(56)

\chi\left(x_{1},x_{2},\ldots,x_{n-1}\right)=\chi(a,a,\ldots,a).

Returning to our problem, we note that, based on formula (44), we have
$\sum_{v=1}^{m-2}(-1)^{v}B_{v}a_{m}^{v}A_{v}^{(n-1)}=(-1)^{m-2}\left[\psi_{m-2}^{(m)}\left(a_{1},a_{2},\ldots,a_{m}\right)-\psi_{m-2}^{(n-1)}\left(a_{1},a_{2},\ldots,a_{m-1}\right)\right]$
Formula (47) therefore gives us

		$\displaystyle\mathrm{A}_{1}^{(m)}\left(a_{1},a_{2},\ldots,a_{m}\right)-(-1)^{m-2}q_{m-2}^{(n)}\left(a_{1},a_{2},\ldots,a_{m}\right)=$
		$\displaystyle\quad=\mathrm{H}_{m-1}\left(a_{1},a_{2},\ldots,\iota_{m-1}\right)-(-1)^{m-2}\psi_{m-2}^{(n-1)}\left(a_{1},a_{2},\ldots,a_{m-1}\right)$

Here the second term depends only on the variables $a_{1},a_{2},\ldots,a_{m-1}$ . and the first term is a symmetric function of $a_{1},a_{2},\ldots,a_{n}$ . Applying Lemma 7 we see that both members reduce to the same constant. However, based on formulas (46), (53), this constant is equal to 0 . It follows that formula (51) for $j=1$ as well as formula (52) are true.
3). Thus: From (55) it is successively deduced

		$\displaystyle\chi\left(x_{1},x_{2},\ldots,x_{m-2},a\right)=\chi\left(x_{1},x_{2},\ldots,x_{m-3},a,a\right)$
		$\displaystyle\chi\left(x_{1},x_{2},\ldots,x_{m-3},a,a\right)=\chi\left(x_{1},x_{2},\ldots,x_{m-4},a,a,a\right)$
		$\displaystyle\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.$
		$\displaystyle\chi\left(x_{1},a,a,\ldots,a\right)=\chi(a,a,\ldots,a)$

do aude formula (56).

With this, Theorem 2 is completely proven.
22. - Formulas (51) allow us to explicitly calculate the polynomial $R_{m}(n)$ We have

\mathrm{R}_{m}(n)=\frac{(-1)^{m-1}}{\delta}\sum_{j=1}^{n-1}\frac{(-1)}{j!}\varphi_{m-1-j}n^{j}

(57)

where it is clear which variables the functions depend on $\varphi_{m-1-j}.\mathrm{Cu}$ using the table of values given in No. 18, the polynomials can be explicitly formed $\mathrm{R}(n)$ for $m=2,3,4,5,6,7,8,9,10,11$ .

§ 5.

23.

–. Let equation (1), to which the abbreviated notations will refer. We will assume that the numbers $a_{1},a_{2},\ldots,a_{m}$ are two primes between them. Let's take the formula (2T) again and put
$\mathrm{M}^{\prime}=\mathrm{M}^{\prime}\left(a_{1},a_{2},\ldots,a_{m}\right)=\sup_{(n)}\mathrm{G}(n),\mathrm{M}^{\prime\prime}=\mathrm{M}^{\prime\prime}\left(a_{1},a_{2},\ldots,a_{m}\right)=\inf_{(n)}(n)$ sup and inf referring to all integer values of $n$ Based on the periodicity of the sequence $\{\mathrm{G}(n)\}$ we have then
$M^{\prime}=\max\mathrm{G}(n)=\max\quad\mathrm{G}(n),M^{\prime\prime}=\min\mathrm{G}(n)=\min\mathrm{G}(n)$
max and min referring to a system of $\delta$ his values $n$ forming a complete system of debris ( $\bmod\delta$ ).

We have
LEMMA 8. - We have the inequality
(58)

M^{\prime}-M^{\prime\prime}\geqq 0

where equality holds if and only if $a_{1}=a_{2}=\ldots=a_{m}=1$ .
Inequality (58) is a consequence of the definition of maximum and minimum. It remains to prove the property relative to equality.

The condition is obviously sufficient, because if it is fulfilled, we have $\delta=1$ and all the numbers $\mathrm{G}(n),n=0,\pm 1,\pm 2,\ldots$ are equal.

The condition is necessary. Indeed, from (18) it also follows

\mathrm{G}(n)=\sum_{i=2}^{m}\mathrm{l}^{4}(1;i)+\mathrm{C}_{1}(n)

from $M^{\prime}=M^{\prime\prime}$ it turns out that the numbers $\mathrm{G}(n)$ , so the numbers $\mathrm{C}_{1}(n),n=0,\pm 1$ . $\pm 2,\ldots$ are all equal. Based on an observation made in No. 12, this is only possible if $a_{1}=a_{2}=\ldots=a_{m}=1$ .

Lemma 8 is completely proven.

24. - We can now demonstrate

THEOREM 3. - The necessary and sufficient condition for problems I, II, II' to have a solution is that we have
(59)

M^{\prime}-M^{\prime\prime}<1

Let us prove the theorem for problem I.
The condition is necessary. Indeed, let $\mathrm{P}(n)$ a solution to the problem. Then

\mathrm{P}(n)=\mathrm{R}(n)+\lambda

(60)

$\lambda$ being 0 constant.
From (4) and (27) we deduce

\mathrm{G}(n)<\lambda+\frac{1}{2},-\mathrm{G}(n)<-\lambda+\frac{1}{2}

whatever $n$ But
there are values $n_{1},n_{2}$ his/hers $n$ as long as $M^{\prime}=G\left(n_{1}\right)M^{\prime\prime}=G\left(n_{2}\right)$ So we have

\mathrm{M}^{\prime}<\lambda+\frac{1}{2},-\mathrm{M}^{\prime\prime}<-\lambda+\frac{1}{2}

Adding these inequalities term by term we find condition (59). The condition is sufficient. Indeed, if inequality (59) is verified, we can find a number $\lambda$ so that
(61)

M^{\prime}-\frac{1}{2}<\lambda<M^{\prime\prime}+\frac{1}{2}

It is then immediately verified that the polynomial (60) verifies the inequality (4). The theorem for problems II, II' is proved in the same way.
If we have (59), all solutions of problem I are contained in the formula
(60) where the constant $\lambda$ check the inequalities (61). Also if we have
(59) all the solutions of problem II are contained in the formula

\text{ (62) }\quad Q(n)=\mathrm{R}(n)+\lambda_{1}

where the constant $\lambda_{1}$ check inequalities

M^{\prime}\leqq\lambda_{1}<M^{\prime\prime}+1

(63)

and all solutions of problem II' are contained in the formula

\mathrm{S}(n)=\mathrm{R}(n)+\lambda_{n}

(64)

where the constant $\lambda_{2}$ check inequalities

M^{\prime}-1<\lambda_{2}\leqq M^{\prime\prime}

(65)

25.

– Any number $\lambda$ which verifies inequalities (61) is of the form

\theta\left(M^{\prime}-\frac{1}{2}\right)+(1-\theta)\left(M^{\prime\prime}+\frac{1}{2}\right)=M^{\prime\prime}+\frac{1}{2}-\theta\left(1-M^{\prime}+M^{\prime\prime}\right)

or in the form of

(1-\theta)\left(M^{\prime}-\frac{1}{2}\right)+\theta\left(M^{\prime\prime}+\frac{1}{2}\right)=M^{\prime}-\frac{1}{2}+\theta\left(1-M^{\prime}+M^{n}\right)

where $\theta$ is an arbitrary positive subunit number.
It follows that the general form of the solutions of problem I is (the polynomial $\mathrm{R}(n)$ being assumed to be normalized by condition (32)),

\mathrm{P}(n)=\mathrm{R}(n)+\mathrm{M}^{\prime\prime}+\frac{1}{2}-\theta\left(1-\mathrm{M}^{\prime}+\mathrm{M}^{\prime\prime}\right);\quad(0<\theta<1)

(66)

\mathrm{P}(n)=\mathrm{R}(n)+\mathrm{M}^{\prime}-\frac{1}{2}+\theta\left(1-\mathrm{M}^{\prime}+\mathrm{M}^{\prime\prime}\right)\quad(0<\theta<1)

(

\gamma

)

It is also seen that the general form of the solutions to problem II is

Q(n)=R(n)+M^{\prime\prime}+1-\theta_{1}\left(1-M^{\prime}+M^{\prime\prime}\right),\quad\left(0<\theta_{1}\leqq 1\right)

(67)

or
( $67^{\prime}$ )

\mathrm{Q}(n)=\mathrm{R}(n)+\mathrm{M}^{\prime}+\theta_{2}\left(1-\mathrm{M}^{\prime}+\mathrm{M}^{\prime\prime}\right),\quad\left(0\leqq\theta_{2}<1\right)

and the general form of the solutions to problem II' is

S(n)=R(n)+M^{\prime}-1+\theta_{1}\left(1-M^{\prime}+M^{\prime\prime}\right),\quad\left(0<\theta_{1}\leqq 1\right)

(68)

or
(68')

S(n)=R(n)+M^{\prime\prime}-\theta_{2}\left(1-M^{\prime}+M^{\prime\prime}\right),\quad\left(0\leqq\theta_{2}<1\right)

It is seen that problems II and II' have exceptional solutions

\mathrm{Q}(n)=\mathrm{R}(n)+\mathrm{M}^{\prime},\quad\mathrm{S}(n)=\mathrm{R}(n)+\mathrm{M}^{\prime\prime}

Apart from these exceptional solutions, the other solutions of problems I, II, II' correspond two by two such that

\mathrm{Q}(n)-\mathrm{P}(n)=\frac{1}{2},\mathrm{P}(n)-\mathrm{S}(n)=\frac{1}{2}\cdot\mathrm{Q}(n)-\mathrm{S}(n)=1

(69)

With this the connection between the solutions of the three problems is completely clarified.
26. -Theorem 3 allows us to easily find criteria for the impossibility of problems I, II II'.

Thus, for example, we have
LEMA 9. - If there are two values $n^{\prime},n^{\prime\prime}$ his/hers $n$ so that

\left|\mathrm{G}\left(n^{\prime}\right)-\mathrm{G}\left(n^{\prime\prime}\right)\right|\geqq 1,

problems I, II, II' have no solutions.
The proof is immediate because under the conditions of the lemma we have

1\leqq\left|\mathrm{G}\left(n^{\prime}\right)-\mathrm{G}\left(n^{\prime\prime}\right)\right|\leqq\mathrm{M}^{\prime}-\mathrm{M}^{\prime\prime}

and it is enough to apply Theorem 3.

We will give several applications of this lemma. For now we will make a first application here. Assuming that the polynomial $\mathrm{R}(n)$ is normalized by condition (30), we always have $\mathrm{N}(0)=1,\mathrm{R}(0)=0$ , so $\mathrm{G}(0)=1$ If the numbers $a_{1},a_{2},\ldots,a_{m}$ they are all $>1$ we obviously have and $\mathrm{N}(1)=0$ , so $\mathrm{G}(1)=-\mathbb{R}(1)$ Therefore $\mathrm{G}(0)-\mathrm{G}(1)=1+\mathrm{R}(1)$ It follows that if $\mathrm{R}(1)\geq 0$ problems I, II, II' have no solution.

But $\mathrm{R}(1)$ is equal to the sum of the coefficients of the polynomial $\mathrm{R}(n)$ and, based on formula (57) and the formulas of No. 18, these coefficients are certainly all positive for $m=1,2,3,4,5$ So we have

THEOREM 4. – If $m=1,2,3,4$ or 5 and if the numbers $a_{1},a_{2}$ , . . , $a_{m}$ they are all $>1$ , problems I, II, II have no solutions.

This property is probably also true for $m>5$ The previous proof is no longer valid, however, because the coefficient

\mathrm{A}_{m-5}^{(m)}=\frac{1}{(m-1)!}q_{4}^{(n)}\left(a_{1},a_{2},\ldots,a_{m}\right)

can also take negative values).

§ 6.

27.
- —
  
  Let's consider the equation

cx=n,

(70)

$c$ being a natural number. We are therefore in the case $m=1$ and then we have $\mathrm{N}(n;c)=1$ or 0 as appropriate $n$ divides or does not divide with $c$ To express the number $N(n;c)$ in a convenient form we will generally denote by $\left(\left.\frac{n}{q}\right\rvert\,p\right)$ the solution, between 0 and $p-1$ , of congruence

qx\equiv n\quad(\bmod p)

—

We will always assume that $p$ and $q$ are prime to each other. Then the number $\left(\left.\frac{n}{q}\right\rvert\,p\right)$ is uniquely and well-determined.

In particular $(n\mid p)$ is equal to the minimum nonnegative remainder of $n(\bmod p)$ We have obviously

\left(\left.\frac{n^{\prime}}{q}\right\rvert\,p\right)=\left(\left.\frac{n^{\prime\prime}}{q}\right\rvert\,p\right),\quad\text{ dacă }n^{\prime}\equiv n^{\prime\prime}(\bmod p)

(71)

We then have
Lemma 10. - The number N( $n;c$ ) is given by the formula

\mathrm{N}(n;c)=\frac{1}{c}+\frac{1}{c}\left\{\left(\left.\frac{n-u}{u}\right\rvert\,c\right)-\left(\left.\frac{n}{u}\right\rvert\,c\right)\right\}

(72)

whatever integer u is prime to c.

⁰⁰footnotetext:

{}^{\text{9 }}

Aceasta nu înseamă că pentru anumite valori particulare ale coeficientilor a/ rationamentul să nu rămână valabile.

Indeed, from

\left(\left.\frac{n-u}{u}\right\rvert\,c\right)u\equiv n-u(\bmod c),\quad\left(\left.\frac{n}{u}\right\rvert\,c\right)u\equiv n\quad(\bmod c),

under the conditions of the lemma, it immediately follows that

\left(\left.\frac{n-u}{u}\right\rvert\,c\right)-\left(\left.\frac{n}{u}\right\rvert\,c\right)=-1\text{ sau }c-1.

The second case occurs if and only if or $c=1$ or $c>1$ and $\left(\left.\frac{n}{u}\right\rvert\,c\right)=0$ , so only if $n$ divides by $c$ .

The lemma follows immediately.
The results established so far apply to equation (70) and problems I, II, II' can be completely solved in this case. The results are good and it is unnecessary to detail them. We have $\mathrm{N}(n;1)=1$ , whatever it is $n$ ,
28. - Let us now consider equation
(73)

bx+cy=n

where $b,c$ are two natural numbers prime between them. We are here in the case $m=2$ Abbreviated notations will refer to equation (73).

We will prove the following
LEMMA 11. - If b, c are prime to each other, we have

\mathrm{N}(n)=\frac{n}{bc}-\frac{1}{b}\left(\left.\frac{n}{c}\right\rvert\,b\right)-\frac{1}{c}\left(\left.\frac{n}{b}\right\rvert\,c\right)+1

(74)

whatever the whole number is $n$
In other words we have

\mathrm{R}(n)=\frac{n}{bc},\mathrm{G}(n)=-\frac{1}{b}\left(\left.\frac{n}{c}\right\rvert\,b\right)-\frac{1}{c}\left(\left.\frac{n}{b}\right\rvert\,c\right)+1.

The first formula was also found in another way.
Based on formulas (24), (72), we have

\mathrm{N}(n+b)-\mathrm{N}(n)=\mathrm{N}(n+b;c)=\frac{1}{c}+\frac{1}{c}\left\{\left(\left.\frac{n}{b}\right\rvert\,c\right)-\left(\left.\frac{n+b}{b}\right\rvert\,c\right)\right\}

If in this formula we successively replace $n$ with $n+b,n+2b,\ldots$ then with $n-b,n-2b,\ldots$ we deduce, adding member by member,

\mathrm{N}(n+\alpha b)-\mathrm{N}(n)=\frac{\alpha}{c}+\frac{1}{c}\left\{\left(\left.\frac{n}{b}\right\rvert\,c\right)-\left(\left.\frac{n+\alpha h}{b}\right\rvert\,c\right)\right\}

(75)

whatever the whole numbers are $n$ and $\alpha$
Based on formula (71), we deduce from (75) ,

\mathrm{N}(n+\alpha b+\beta c)-\mathrm{N}(n+\beta c)=\mathrm{N}(n+\alpha b)-\mathrm{N}(n)

whatever the integers are $n,\alpha,\beta$ . Doing $n=0$ we deduce from here
(76) $N(\alpha b+\beta c)=[N(\alpha b)-N(0)]+[N(\beta c)-N(0)]+N(0)$ ,

But from (75) we deduce, making $n=0$ ,

\mathrm{N}(\alpha,b)-\mathrm{N}(0)=\frac{\alpha}{c}-\frac{1}{c}(x\mid c)

(77)

Analogously we have

\mathrm{N}(\beta c)-\mathrm{N}(0)=\frac{\beta}{b}-\frac{1}{b}(\beta\mid b)

(78)

Be it now $n$ an arbitrary integer and determine the integers $\alpha,\beta$ so that we have

n=\alpha b+\beta c

Then from $n\equiv\alpha b(\bmod c),n\equiv\beta c(\bmod b)$ it results that $(\alpha,c)==\left(\left.\frac{n}{b}\right\rvert\,c\right),(\beta\mid b)=\left(\left.\frac{n}{c}\right\rvert\,b\right)$ .

Taking into account $\mathrm{N}(0)=1$ and from formulas (76), (77), (78), we deduce formula (74).

Lemma 11 is therefore proven.
29. - In the case of equation (73) we can easily calculate the numbers $\mathrm{M}^{\prime},\mathrm{M}^{\prime\prime}$ We have

\mathrm{M}^{\prime}=1,\mathrm{M}^{\prime\prime}=\frac{1}{b}+\frac{1}{c}-1

the maximum being reached when $n\equiv 0(\bmod bc)$ , and the minimum when $n\equiv-b-c(\bmod bc)$ .

It is immediately verified that for inequality (59) to be satisfied it is necessary and sufficient that at least one of the numbers $b,c$ to be equal to $1.{}^{7}$ )

So be it. $b=1$ We then have $\mathrm{R}(n)=\frac{n}{c}$ and $\mathrm{M}^{\prime}-\mathrm{M}^{\prime\prime}=\frac{c-1}{c}<1$ so the results of the previous § allow us to state

THEOREM 5. - Number $N(n)$ relative to the equation $x+cy=n$ is equal to whatever integer n is, with:
I. The integer closest to the number $\frac{2n+c+2\theta}{2c},(0<\theta<1)$ .
II. The whole contained in the number $\frac{n+c+\theta_{2}}{c},\left(0\leqq\theta_{2}<1\right)$ .

II'. The integer that contains the number $\frac{n+\theta_{1}}{c},\left(0<\theta_{1}\leqq 1\right)$ .

⁰⁰footnotetext: 7. Inegalitatea (59) révine la

1<\frac{1}{b}+\frac{1}{c}

care úu poate fi satisfăcută pentru

b,o\geqq 2

A simple particular statement is the following:
The number $\mathrm{N}(n)$ relative to the equation $x+cy=n$ is equal to the integer that contains the number $\frac{2n+1}{2c}$ , whatever the integer n.
30. - In the case $m=2$ we can easily directly demonstrate the necessity of the condition of Theorem 1. Suppose that in equation (73) $b$ and $c$ are any two natural numbers. Let $d$ cmmdc of these numbers. Then $b=db_{1},c=dc_{1}$ , where $b_{1},c_{1}$ are two natural numbers that are prime to each other. If we assume $d>1$ HAVE

\mathrm{N}(dn+1)=0,\mathrm{\penalty 10000\ N}(dn)=\mathrm{N}\left(n;b_{1},c_{1}\right)

(79)

whatever the whole number is $n$
The first equality results from the fact that $bx+cy$ divides by $d$ whatever the whole numbers are $x,y$ , and $dn+1$ is prime with $d$ The second equality (79) results from the fact that any solution of the equation $bx+cy=d\left(b_{1}x+c_{1}y\right)==dn$ is a solution to the equation $b_{1}x+c_{1}y=n$ and reciprocally.

Let us now assume that problem III has a solution R(n).
The first formula (79) then shows us that

\mathrm{R}(dn+1)=\lambda

(80)

$\lambda$ being a constant and whatever it is $n$
The second formula (79), together with (74), shows us that
$\lambda^{\prime}$ being a constant, whatever it is $n$ .
But from (80) it follows that $\mathrm{R}(n)=\lambda$ and from (81) it follows that $\mathrm{R}(n)==\frac{n}{db_{1}c_{1}}+\lambda^{\prime}$ , whatever it is $n$ We should have

\lambda=\frac{n}{db_{1}c_{1}}+\lambda^{\prime}

whatever $n$ , which is clearly impossible. The property is therefore proven.

§7.

31.
- —
  
  Let us now consider the case $m=3$ and therefore let equation
  (82) be

ax+by+cz=n

$a,b,c$ , being three natural numbers two by two prime to each other. The abbreviated notations will now refer to this equation.

To express the number $N(n)$ we will introduce the following function

(n;q\mid p)=\frac{1}{p^{2}}\sum_{i=0}^{n-1}i(n-qi\mid p)

(83)

where $p,q$ are two natural numbers that are prime to each other.

It is easy to see that we have
(84) $\left(n^{\prime};q^{\prime}|p\rangle=\left(n^{\prime\prime};q^{\prime\prime}\mid p\right)\right.$ , if $n^{\prime}\equiv n^{\prime\prime},q^{\prime}\equiv q^{\prime\prime}(\bmod p)$

We also have the formula

(n+1;q\mid p)-(n;q\mid p)=\frac{p-1}{2p}-\frac{1}{n}\left(\left.\frac{n+1}{q}\right\rvert\,p\right)

(85)

whose proof is made by observing that the first member is equal to

\frac{1}{p^{2}}\sum_{i=0}^{p-1}i[(n+1-qi\mid p)-(n-qi\mid p)]

and that we have

(n+1-qi\mid p)-(n-qi\mid p)=\begin{cases}1,&\text{ dacă }qi\neq n+1\quad(\bmod p)\\ -p+1,&\text{ dacă }qi\equiv n+1\quad(\bmod p)\end{cases}

because the numbers $n-qi,i=0,1,\ldots,p-1$ forms a complete system of debris ( $\bmod p$ ).

Formula (83) can also be written

(n;q\mid p)=\frac{1}{x^{2}}\sum_{i=0}^{n-1}i\left(nq^{\prime}-iq^{\prime}\mid p\right)

where $q^{\prime}=\left(\left.\frac{1}{q}\right\rvert\,p\right)$ This follows from the fact that if $n-qj\equiv i(\bmod p)$ we will also have $j\equiv nq^{\prime}-iq^{\prime}(\bmod p)$ .

It follows that we have
(86) $\quad(n;q\mid p)=\left(n_{1};q^{\prime}\mid p\right)$ , if $qq^{\prime}\equiv 1,n_{1}\equiv nq^{\prime}\quad(\bmod p)$ .

Finally, we also observe that

(n;q;1)=0

whatever $n$ and $q$ .
32. - We will now demonstrate

LEMMA 12, - If $a,b,c$ are two prime numbers between them, the number $\mathrm{N}(n)$ , corresponding to equation (82), is given by formula
(88)
where
(89)
(90)

\begin{gathered}\mathrm{R}(n)=\frac{n(n+a+b+c)}{2abc}\\ \mathrm{G}(n)=\left(nc^{\prime};c^{\prime}b\mid a\right)-\left(0;c^{\prime}b\mid a\right)+\left(na^{\prime};a^{\prime}c\mid b\right)-\left(0;a^{\prime}c\mid b\right)+\\ +\left(nb^{\prime};b^{\prime}a\mid c\right)-\left(0;b^{\prime}a\mid c\right)+1\\ \left.a^{\prime}=\left(\left.\frac{1}{a}\right\rvert\,b\right),b^{\prime}=\left(\left.\frac{1}{b}\right\rvert\,c\right),c^{\prime}=\left(\left.\frac{1}{c}\right\rvert\,a\right)^{8}\right)\end{gathered}

⁰⁰footnotetext: 8. Nu respective.

For demonstration we will also put

a^{\prime\prime}=\left(\left.\frac{1}{a}\right\rvert\,c\right)\cdot b^{\prime\prime}=\left(\left.\frac{1}{b}\right\rvert\,a\right),c^{\prime\prime}=\left(\left.\frac{1}{c}\right\rvert\,b\right)

Pe baza formulator (24), (74), avem
$\mathrm{N}(n+c)-\mathrm{N}(n)=\mathrm{N}(n+c;a,b)=\frac{n+c}{ab}-\frac{1}{a}\left(\frac{n+c}{b}\left\lvert\,a\left(-\frac{1}{b}\left(\left.\frac{n+c}{a}\right\rvert\,b\right)+1\right.\right.\right.$
But

\left(\left.\frac{n+c}{b}\right\rvert\,a\right)=\left(\left.\frac{c^{\prime}n+1}{c^{\prime}b}\right\rvert\,a\right),\left(\left.\frac{n+c}{a}\right\rvert\,b\right)=\left(\left.\frac{c^{\prime\prime}n+1}{c^{\prime\prime}a}\right\rvert\,b\right)

and taking into account formula (85) we then deduce

\mathrm{N}(n+c)-\mathrm{N}(n)=\frac{2n+2c+a+b}{2ab}+\left(c^{\prime\prime}n+1;c^{\prime}b\mid a\right)-\left(c^{\prime}n;c^{\prime}b\mid a\right)

If in this equality we replace $n$ successively with $n,n+c.n+2c$ ; $\ldots$ and then successively with $n-c,n-2c,\ldots$ , if we observe that $i$ being an arbitrary integer,

c^{\prime}(n+ic)\equiv c^{\prime}n+i(\bmod a),c^{\prime\prime}(n+ic)\equiv c^{\prime\prime}n+i(\bmod b)

and finally if we add the equalities thus obtained member by member, we deduce

\begin{gathered}\mathrm{N}(n+\mu c)-\mathrm{N}(n)=\mu\frac{2n+\mu c+a+b+c}{2ab}+\left(c^{\prime}n+\mu;c^{\prime}b\mid a\right)-\\ -\left(c^{\prime}n;c^{\prime}b\mid a\right)+\left(c^{\prime\prime}n+\mu;c^{\prime\prime}a\mid b\right)-\left(c^{\prime\prime}n;c^{\prime\prime}a\mid b\right)\end{gathered}

whatever the integers are $n$ and $\mu$ Putting
$\mu=\alpha b$ , where $\alpha$ is an integer and taking into account (84), we deduce

\mathrm{N}(n+\alpha bc)-\mathrm{N}(n)=\alpha\frac{2n+\alpha bc+a+b+e}{2a}+\left(c^{\prime}n+ab;c^{\prime}b\mid a\right)-

(91)

whatever the numbers are $n$ and $\alpha$ If
in this formula we replace $n$ with $n+\beta ac$ and taking into account (84) again, we deduce
(92) $\mathrm{N}(n+\alpha bc+\beta ca)-\mathrm{N}(n+\alpha bc)-\mathrm{N}(n+\beta ca)+\mathrm{N}(n)=\alpha\beta c$ , whatever the integers are $n,\alpha.\beta$ .

Doing here $n=0$ , we obtain
(93) $\mathrm{N}(\alpha bc+\beta ca)-\mathrm{N}(\alpha bc)-\mathrm{N}(\beta ca)+\mathrm{N}(0)=\alpha\beta c$ and analogously we deduce
(93') $\left\{\begin{array}[]{l}\mathrm{N}(\beta ca+\gamma ab)-\mathrm{N}(\beta ca)-\mathrm{N}(\gamma ab)+\mathrm{N}(0)=\beta\gamma a,\\ \mathrm{\penalty 10000\ N}(\gamma ab+\alpha bc)-\mathrm{N}(\gamma ab)-\mathrm{N}(\alpha bc)+\mathrm{N}(0)=\gamma\alpha b,\end{array}\right.$
whatever the whole numbers are $\alpha,\beta,\gamma$ .

The first member of formula (92) being independent of $n$ we deduce, equating the values of this expression for $n=0$ and $n=\gamma ab$ ,
(94) $\mathrm{N}(\alpha bc+\beta ca+\gamma ab)=[\mathrm{N}(\beta ca+\gamma ab)-\mathrm{N}(\beta ca)-\mathrm{N}(\gamma ab)+\mathrm{N}(0)]+$

\begin{gathered}+[\mathrm{N}(\gamma ab+\alpha bc)-\mathrm{N}(\gamma ab)-\mathrm{N}(\alpha bc)+\mathrm{N}(0)]+\\ +[\mathrm{N}(\alpha bc+\beta ca)-\mathrm{N}(\alpha bc)-\mathrm{N}(\beta ca)+\mathrm{N}(0)]+\\ +[\mathrm{N}(\alpha bc)-\mathrm{N}(0)]+[\mathrm{N}(\beta ca)-\mathrm{N}(0)]+[\mathrm{N}(\gamma ab)-\mathrm{N}(0)]+\mathrm{N}(0).\end{gathered}

From (91), making $n=0$ , we deduce
(95) $\mathrm{N}(\alpha bc)-\mathrm{N}(0)=\alpha\frac{\alpha bc+a+b+c}{2a}+\left(\alpha b;c^{\prime}b\mid a\right)-\left(0;c^{\prime}b\mid a\right)$
and similarly we obtain
$\left(95^{\prime}\right)\left\{\begin{array}[]{l}\mathrm{N}(\beta ca)-\mathrm{N}(0)=\beta\frac{\beta ca+a+b+c}{2b}+\left(\beta c;a^{\prime}c\mid b\right)-\left(0;a^{\prime}c\mid b\right),\\ \mathrm{N}(\gamma ab)-\mathrm{N}(0)=\gamma\frac{\gamma ab+a+b+c}{2c}+\left(\gamma a;b^{\prime}a\mid c\right)-\left(0;b^{\prime}a\mid c\right).\end{array}\right.$
Considering the formulas $(93),\left(93^{\prime}\right),(95),\left(95^{\prime}\right)$ and noting that $\mathrm{N}(0)=1$ , formula (94) becomes
$\mathrm{N}(\alpha bc+\beta ca+\gamma ab)=\frac{(\alpha bc+\beta ca+\gamma ab)^{2}+(a+b+c)(\alpha bc+\beta ca+\gamma ab)}{2abc}+$
$+\left(\alpha b;c^{\prime}b\mid a\right)-\left(0;c^{\prime}b\mid a\right)+\left(c;a^{\prime}c\mid b\right)-\left(0;a^{\prime}c\mid b\right)+$
$\left(\gamma a;b^{\prime}a\mid c\right)-\left(0;b^{\prime}a\mid c\right)+1$
But $n$ being an integer, and the numbers $bc,ca,ab$ , being prime among themselves, we can always find the integers $\alpha,\beta,\gamma$ so that

\alpha bc+\beta ca+\gamma ab=n.

We then have

\alpha b\equiv nc^{\prime}(\bmod a),\beta c\equiv na^{\prime}(\bmod b),\gamma a\equiv nb^{\prime}(\bmod c)

and it is immediately seen that Lemma 12 follows.
If $a=1$ , we also have $a^{\prime}=a^{\prime\prime}=1$ If we also observe that based on formula (86) we have

\left(nb^{\prime};b^{\prime}a\mid c\right)=\left(na^{\prime\prime};a^{\prime\prime}b\mid c\right)

deduce

\mathbf{R}(n;1,b,c)=\frac{n(n+b+c+1)}{2bc}

$\mathrm{G}(n;1,b,c)=(n;c\mid b)-(0;c\mid b)+(n;b\mid c)-(0;b\mid c)+1$
Finally for $a=b=1$ we will have

		$\displaystyle\mathrm{R}(n;1,c)=\frac{n(n+c+2)}{2c}$
		$\displaystyle\mathrm{G}(n;1,c)=(n;1\mid c)-(0;1\mid c)+1$

33.
- —
  
  Before going further, we will establish the following property of sums (83):

LEMMA 13. - We have the formula

(-n-1-q;q\mid p)=(n;q\mid p)

whatever the integer n is.
To prove the lemma let us put

\mathrm{T}_{n}=(n;q\mid p)-(-n-1-q;q\mid p).

Based on formula (85) we then have

\mathrm{T}_{n+1}-\mathrm{T}_{n}=\frac{p-1}{p}-\frac{1}{p}\left[\left(\left.\frac{n+1}{q}\right\rvert\,p\right)+\left(\left.\frac{-n-1-q}{q}\right\rvert\,p\right)\right]

But from

\left(\left.\frac{n+1}{q}\right\rvert\,p\right)q\equiv n+1,\left(\left.\frac{-n-1-q}{q}\right\rvert\,p\right)q\equiv-n-1-q(\bmod p)

deduce

\left(\left.\frac{n+1}{q}\right\rvert\,p\right)+\left(\left.\frac{-n-1-q}{q}\right\rvert\,p\right)\equiv-1(\bmod p)

from where

\left(\left.\frac{n+1}{q}\right\rvert\,p\right)+\left(\left.\frac{-n-1-q}{q}\right\rvert\,p\right)=p-1.

We therefore have $\mathrm{T}_{n+1}=\mathrm{T}_{n}$ whatever $n$ and it follows from this that (96)

\mathrm{T}_{n}=\mathrm{T}_{0}

We will now show that
(97)

\mathrm{T}_{0}=0

For this we have, taking into account (85),

\begin{gathered}\mathrm{T}_{0}=(0;q\mid p)-(-1-q;q\mid p)=(0;q\mid p)-(-q;q\mid p)+[(-q;q\mid p)-\\ -(-1-q;q\mid p)]=(0;q\mid p)-(-q;q\mid p)+\frac{p-1}{2p}-\frac{1}{p}\left(\left.\frac{-q}{q}\right\rvert\,p\right)=\\ \quad=(0;q\mid p)-(-q;q\mid p)-\frac{p-1}{2p}\end{gathered}

	$\displaystyle(0;q\mid p)-(-q;q\mid p)$	$\displaystyle=\frac{1}{p^{2}}\sum_{i=0}^{p-1}i(-qi\mid p)-\frac{1}{p^{2}}\sum_{i=0}^{p-1}i(-q(i+1)\mid p)=$
		$\displaystyle=\frac{1}{p^{2}}\sum_{i=u}^{p-1}(-qi\mid p)=\frac{p-1}{2p}$

for $-qi,i=0,1,\ldots,p-1$ forms a complete waste system $(\bmod p)$ .

Formula (97) results, and from (96) it then follows that $\mathrm{T}_{n}=0$ , whatever it is $n$ and Lemma 13 is proven.
34. - Let's return to our problem. We will prove the following

THEOREM 6. - The function G(n), periodic and of period abc, is symmetric about the center of symmetry - $\frac{a+b+c^{9}}{2}$ ).

Of course the function $\mathrm{G}(n)$ has, due to periodicity, an infinity of symmetry centers.

The theorem is expressed by the equality

\mathrm{Q}(-a-b-c+1-k)=\mathrm{Q}(k-1),

whatever the whole number is $k$ , or by other analogous equalities.
Based on formula (90) it is enough to prove the property for each of the functions
(98)

\left(nc^{\prime};c^{\prime}b\mid a\right),\quad\left(na^{\prime};a^{\prime}c\mid b\right),\quad\left(nb^{\prime};b^{\prime}a\mid c\right).

For the first of these functions the property returns to the equality

\left(c^{\prime}(k-1);c^{\prime}b\mid a\right)=\left(-c^{\prime}(k-1)-1-c^{\prime}b;c^{\prime}b\mid a\right)

which is true based on Lemma 13. The property for the other two functions (98) is proven in the same way.

Theorem 6 is therefore proven.
35. The previous results can still be specified.

If a function is periodic by period $\delta$ and if it has a center of symmetry $n_{0}$ , any point congruent with $n_{0}\left(\bmod\frac{\delta}{2}\right)$ is a center of symmetry.

It follows from Theorem 6 that for the function $\mathrm{G}(\hat{\imath})$ point $y=\frac{abc-a-b-c}{2}$ is a center of symmetry. The numbers $a,b,c$ being two primes between them, at most one of them is even so that the number $v$ is always whole.

It turns out that we have

\mathrm{G}(v+k)=\mathrm{G}(v-k)

whatever the whole number is $k$ .
Formula (89) however gives us

\mathrm{R}(\nu+k)-\mathrm{R}(\nu-k)=k.

⁰⁰footnotetext: ⁹⁾ funcția

f(x)

este simetrică faţă de centrul de simetrie vo dacă revem, pentru orice

x

f\left(x_{0}+x\right)=f\left(x_{0}-x\right)

So we have
(99)

\mathrm{N}(v+k)-\mathrm{N}(v-k)=k.

whatever the whole number is $k$ If
$n$ is negative and $>-a-b-c$ HAVE $\mathrm{N}(n)=0$ , as follows from the definition of the number $N(n)$ . From (99) it follows that

\mathrm{N}(2v+i)=v+i,\quad i=1,2,\ldots,a+b+c-1.

Taking into account formulas (88) and (89) and doing the calculations we find

\mathrm{G}(2\nu+i)=\frac{i(a+b+c-i)}{2abc},i=1,2,\ldots,a+b+c-1.

(100)

row

\mathrm{G}(0),\mathrm{G}(1),\ldots,\mathrm{G}(abc-1)

which forms a period of the sequence $\{G(n)\}$ so it is divided into two sections

\mathrm{G}(0),\cdot\mathrm{G}(1),\ldots,\mathrm{G}(2v),

(101)

\mathrm{G}(2v+1),\mathrm{G}(2v+2),\ldots,\mathrm{G}(abc-1)

(102)

each being symmetric in the sense that terms equidistant from the extremes are equal. Furthermore, formula (100) shows us that the sequence (102) is constituted by the values of a polynomial of the second degree (its third difference is zero).

In the particular case $2v<0$ , which occurs if and only if at least two of the numbers $a,b,c$ are equal to $1^{10}$ ), the string (101) disappears and only the symmetric string (102) remains. In this case, by the way $y=-1$ .
36. - To conclude this § we will give a direct demonstration of the necessity of the condition in Theorem 1 in the particular case $m=3$ .

We will now assume that in equation (82) $a,b,c$ are three arbitrary natural numbers.

It is known that the numbers $a,b,c$ , can always be written in the form [6]

a=db_{1}c_{1}a_{2},\quad b=dc_{1}a_{1}b_{2},\quad c=da_{1}b_{1}c_{2},

where $d,a_{1},b_{1},c_{1},a_{2},b_{2},c_{2}$ there are 7 natural numbers such that:
$1^{\circ}a_{1},b_{1},c_{1}$ there are two prime numbers between them,
$2^{\circ}a_{2},b_{2},c_{2}$ there are two primes between them,
between them the pairs $a_{1},a_{2};b_{1},b_{2};c_{1},c_{2}$ are groups of two prime numbers
I
respect $d$ is the gcdc of numbers $a,b,c$ and $dc_{1},da_{1},db_{1}$
iv cmmde of groups of two numbers $a,b;b,c;c,a$ .
¹⁰⁾ Inequality $2\nu<0$ returns to (*) $1<\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}$ If at least two numbers are $a,b,c$ are $>1$ we have, conveniently fixing the notations, $a\geqq 1,b\geqq 2,c\geqq 3$ , from where $\frac{1}{ab}+\frac{1}{a}+\frac{1}{a^{6}}\leqslant 1$ , which contradicts the inequality (%).

The necessary and sufficient condition that the numbers $a,b,c$ to be two primes between each other is to have $d=a_{1}=b_{1}=c_{1}=1$ .

Let's assume that this condition is not met but that
there is a solution $\mathrm{R}(n)$ of problem III. We then distinguish two cases: $1^{\circ}.d>1$ In this case, as in No. 30, it is seen that

\mathrm{N}\left(da_{1}b_{1}c_{1}n+1\right)=0

(103)

—

whatever $n$ .

But we also have
(104)

\mathrm{N}\left(da_{1}b_{1}c_{1}n\right)=\mathrm{N}\left(n;a_{2},b_{2},c_{2}\right).

Indeed, any solution to the equation

ax+by+cz=d\left(b_{1}c_{1}a_{2}x+c_{1}a_{1}b_{2}y+a_{1}b_{1}c_{2}z\right)=da_{1}b_{1}c_{1}n

is a solution to the equation

b_{1}c_{1}a_{2}x+c_{1}a_{1}b_{2}y+a_{1}b_{1}c_{2}z=a_{1}b_{1}c_{1}n

(105)

and vice versa. Avernus therefore
(106)

\mathrm{N}\left(da_{1}b_{1}c_{1}n\right)=\mathrm{N}\left(a_{1}b_{1}c_{1}n;b_{1}c_{1}a_{2},c_{1}a_{1}b_{2},a_{1}b_{1}c_{2}\right).

But if we have (105), the integers $x,y,z$ are respectively divisible by $a_{1},b_{1},c_{1}$ , based on the reported properties of numbers $a_{1},b_{1},c_{1},a_{2},b_{2},c_{2}$ It follows that the solutions of the equation

b_{1}c_{1}a_{2}x+c_{1}a_{1}b_{2}y+a_{1}b_{1}c_{2}z=a_{1}b_{1}c_{1}\left(a_{2}x^{\prime}+b_{2}y^{\prime}+c_{2}z^{\prime}\right)=a_{1}b_{1}c_{1}n

correspond one-to-one with the solutions of the equation $a_{2}x+b_{2}y+c_{2}z=n$ We have by reinforcement
(107)

\mathrm{N}\left(a_{1}b_{1}c_{1}n;b_{1}c_{1}a_{2},c_{1}a_{1}b_{2},a_{1}b_{1}c_{2}\right)=\mathrm{N}\left(n;a_{2},b_{2},c_{2}\right)

Formula (104) results from (106) and (107).
From formulas (103) and (104) we deduce respectively

	$\displaystyle\mathrm{R}\left(da_{1}b_{1}c_{1}n+1\right)=\lambda$		(108)
	$\displaystyle\mathrm{R}\left(da_{1}b_{1}c_{1}n\right)=\frac{n\left(n+a_{2}+b_{2}+c_{2}\right)}{2a_{2}b_{2}c_{2}}+\lambda^{\prime}$		(109)

whatever $n$ , and $\lambda,\lambda^{\prime}$ being constants.
But from (108) it follows $\mathrm{R}(n)=\lambda$ and from (109) it follows $\mathrm{R}(n)==\frac{dn\left(n+aa_{1}+bb_{1}+cc_{1}\right)}{2abc}+\lambda^{\prime}$ . So we should have

\lambda=\frac{an\left(n+aa_{1}+bb_{1}+cc_{1}\right)}{2abc}+\lambda^{\prime}

whatever it is of what is visibly impossible.
20. $d=1$ In this case at least one of the mummers $a_{1},b_{1},c_{1}$
is $>1$ Either, for fixing the notations, $c_{1}>1$ To simplify the discussion we can assume $n$ positive.

A reasoning analogous to the one above shows us that

\mathrm{N}\left(a_{1}b_{1}c_{1}n-1,a_{1}b_{1}c_{2}\right)=\mathrm{N}\left(nc_{1}+c_{2};c_{1}a_{2},c_{1}b_{2},c_{2}\right)

But if we have

c_{1}a_{2}x+c_{1}b_{2}y+c_{2}z=c_{1}n+c_{2}

it must be $z-1$ to divide with $c_{1}$ , because $c_{1},c_{2}$ are prime to each other.
Suppose that

z=1+c_{1}t

(110)

then

a_{2}x+b_{2}y+c_{2}t=n

Because $c_{1}>1$ , from (110) it is seen that the numbers $z,t$ are always both positive or zero. It follows by simple reasoning that

\mathrm{N}\left(nc_{1}+c_{2};c_{1}a_{2},c_{1}b_{2},c_{2}\right)=\mathrm{N}\left(n;a_{2},b_{2},c_{2}\right)

So finally
(111)

\mathrm{N}\left(a_{1}b_{1}c_{1}n+a_{1}b_{1}c_{2}\right)=\mathrm{N}\left(n;a_{2},b_{2},c_{2}\right)

From formulas (104) ( $d=1$ ) and (111) we deduce

		$\displaystyle\mathrm{R}\left(a_{1}b_{1}c_{1}n+a_{1}b_{1}c_{2}\right)=\frac{n\left(n+a_{2}+b_{2}+c_{2}\right)}{2a_{2}b_{2}c_{2}}+\lambda$
		$\displaystyle\mathrm{R}\left(a_{1}b_{1}c_{1}n\right)=\frac{n\left(n+a_{2}+b_{2}+c_{2}\right)}{2a_{2}b_{3}c_{2}}+\lambda^{\prime}$

		$\displaystyle\mathrm{R}(n)=\frac{(n-c)\left(n-c+aa_{1}+bb_{1}+cc_{1}\right)}{2abc}+\lambda$
		$\displaystyle\mathrm{R}(n)=\frac{n\left(n+aa_{1}+bb_{1}+cc_{1}\right)}{2abc}+\lambda^{\prime}$

therefore

\frac{n\left(n+aa_{1}+bb_{1}+cc_{1}\right)}{2abc}+\lambda^{\prime}=\frac{(n-c)\left(n-c+aa_{1}+bb_{1}+cc_{1}\right)}{2abc}+\lambda

whatever $n$ and $\lambda;\lambda^{\prime}$ being constants. It is easy to see that this is impossible.

With this the intended property is completely demonstrated.

§ 8.

37.
- —
  
  We will now deal with solving problems I, II, II' in the case $m=3$ We have seen that it is sufficient to assume that one of the coefficients is equal to 1. So let equation
  (112) be

x+by+cz=n

where, as we have seen, we can assume the numbers $b,c$ primes between them. For the purpose of fixing the notations we will assume $b\leq c$ . Then we have or $b=c=1$ or $t<c$ . The abbreviated notations will now refer to equation (112) _of
38. - Let us first assume that $b=1$ Based on what has been established above, we then have

M^{\prime}=\max_{i=0,1,\ldots,c-}

The calculation of M' and M” is easy by observing that G (n) for the considered values of $n$ is a second degree polynomial and that a second degree polynomial is a convex or concave function symmetric with respect to its relative extremum point as the center of symmetry. Such a calculation will be repeated several times in the following.

Doing the calculations we find

		$\displaystyle\mathbf{M}^{\prime}=$
		$\displaystyle\mathbf{M}^{\prime\prime}=\frac{c+1}{2c}$

M^{\prime}-M^{\prime\prime}=\begin{cases}\frac{c}{8},&\text{ pentru }c\text{ par }\\ \frac{c^{2}-1}{8c},&\text{ pentru }c\text{ impar }\end{cases}

It is seen that problems I, II, II' have solutions if and only if $c=1,2,3,4,5,6,7$ .

The final result will be stated below.
39. - Let us now suppose that $b>1$ . We then necessarily have $b<c$ and $\nu\geqq 0$ .

If we put

$\mathrm{M}_{1}^{\prime}=\max_{i=0,1,\ldots,v}\{\mathrm{G}(i)\}$ ,	$\mathrm{M}_{1}^{\prime\prime}=$	$\min_{i=0,1,\ldots,v}\{\mathrm{G}(i)\}$
$\mathrm{M}_{2}^{\prime}=\max_{i=1,2,\ldots,b+c}\{\mathrm{G}(2v+i)\}$	$\mathrm{M}_{2}^{\prime\prime}=\min_{i=1,2,\ldots,b+c}\{\mathrm{G}(2v+i)\}$

we will have
(113)

M^{\prime}=\max\left\{M_{1}^{\prime}\cdot M_{2}^{\prime}\right\},\quad M^{\prime\prime}=\min\left\{M_{1}^{\prime\prime},M_{2}^{\prime\prime}\right\}

We thus obtained some formulas that will serve to effectively calculate the extremes $\mathrm{M}^{\prime},\mathrm{M}^{\prime\prime}$ .

The numbers $M_{2}^{\prime},M_{2}^{\prime\prime}$ is calculated as in No. 38, using formula (100), in which we put $a=1$ .

A simple calculation gives us

M_{2}^{\prime}=\begin{cases}\frac{(b+c+1)^{2}}{8bc},&\text{ dacă }b,c\text{ sunt de paritate diferită, }\\ \frac{(b+c)(b+c+2)}{8bc},&\text{ dacă }b,c\text{ sunt ambele impare }\\ M_{2}^{\prime\prime}=\frac{b+c}{2bc}&\end{cases}

Calculating numbers $M_{1}^{\prime},M_{1}^{\prime\prime}$ can be simplified based on the following LEMMA 14. - If $0\leqq n<bc$ , we have $0\leqq\mathrm{\penalty 10000\ N}(n;b,c)\leqq 1$ .
The first inequality is obvious. It is therefore sufficient to prove the second inequality. This inequality means that the equation $bx+cy=n$ , for $0\leqq n<bc$ has at most one solution in nonnegative integers. Suppose the opposite and then let $x,y;x^{\prime},y^{\prime}$ two solutions of this equation. We have $bx+cy=bx^{\prime}+cy^{\prime}<bc$ from where

\begin{array}[]{ll}0\leqq x,x^{\prime}<c,&0\leqq y,y^{\prime}<b\\ x\equiv x^{\prime}(\bmod c),&y\equiv y^{\prime}\quad(\bmod b)\end{array}

which attracts $x=x^{\prime},y=y^{\prime}$ .
Lemma 14 is therefore proven.
If we now take into account:
$1^{\circ}$ Inequality $v<bc-1$ ,
$2^{\circ}$ . Lemma 14,
$3^{\circ}$ . Formula
(115)

\mathrm{N}(n+1)-\mathrm{N}(n)=\mathrm{N}(n+1;b,c)

4.

The fact that the polynomial $\mathrm{R}(n)=\frac{n(n+b+c+1)}{2bc}$ is growing for $n\geqq 0$ ,
we immediately find
(116) $\mathrm{M}_{1}^{\prime}=\max_{o\leqq bx+cy\leqq\nu}\{\mathrm{G}(bx+cy)\},\mathrm{M}_{1}^{\prime\prime}=\min_{o\leqq bx+cy-1\leqq\nu}\{\mathrm{G}(bx+cy-1)\}$ where $x,y$ goes through all possible non-negative integer values.

In applications. $\mathrm{G}(n)$ is calculated from the formula $\mathrm{G}(n)=\mathrm{N}(n)-\mathrm{R}(n)$ , directly evaluating the $\mathrm{N}(n)$ . We will make such applications below.
40. - Let $s$ how much and $r$ the remainder of his division $c$ by $b$ . So $c=sb+r$ , $0<s,0<r<b$ and $r$ is prime with $b$ .

If $0\leqq n<c$ , equation (112) can only be satisfied for $z=0$ , and the unknown $y$ then take the values $\left.0,1,\ldots,\frac{n}{b}\right\rfloor$ . Result - therefore that

N(n;=N(n:1,b),\operatorname{dac}a0\leqq n<c.

But
(117)
whatever it is $n$ .

\mathrm{N}(n;1,b)=\left[\frac{n}{v}\right]+1

We therefore have

\mathrm{N}(n)=\left\lceil\frac{n}{b}\right\rceil+1,\text{ dacă }0\leqq n<c

(118)

From this formula it follows that $\mathrm{N}((s-1)b)=\mathrm{N}(sb-1)=s$ and an elementary calculation shows us that

\begin{gathered}\mathrm{M}^{\prime}-\mathrm{M}^{\prime\prime}\geqq\mathrm{G}((s-1)b)-\mathrm{G}(sb-1)=\mathrm{R}(sb-1)-\mathrm{R}((s-1)\mathrm{b})=\\ =-\frac{3sb(b-1)+r(b-1)}{2b(sb+r)}\end{gathered}

Problems I, II, II' will have no solution if this ratio is $\geq 1.0$ A necessary condition for these problems to have a solution is therefore that this report be $<1$ , condition that can be written in the form. s $b(b-3)-r(b+1)<0$ or

\frac{sb(b-3)}{b+1}<r

(119))

To find another delimitation let's take $n=(s+1)b$ In this case equation (112) cannot be satisfied unless $z=0$ or 1. It follows, taking into account (117), that

\mathrm{N}((s+1)b)=\mathrm{N}((s+1)b;1,b)+\mathrm{N}(b-r;1,b)=s+3

We then have

\mathrm{M}^{\prime}\geqq\mathrm{G}((s+1)b)=s+3-\frac{(s+1)b[2(s+1)b+r+1]}{2b(sb+r)}

On the other hand, based on formulas (113) and (114), we have

It follows that

\mathrm{M}^{\prime\prime}\leqq\mathrm{M}_{2}^{\prime\prime}=\frac{(s+1)b+r}{2b(sb+r)}

M^{\prime}-M^{\prime\prime}\geq\frac{2b[(s-1)b-s+1]+r[(s+5)b-1]}{2b(sb+r)}

from which, proceeding as above, a second necessary condition for the possibility of problems I, II, II' is deduced, a condition which is written

-2b(b+s+1)+r[(s+3)b-1]<0

r<\frac{2b(b+s+1)}{(s+3)b-1}

(120)

From incrality (119), (120); having regard to the values peculiar to
him $b,r$ and $s$ its values are deduced $b,c$ which satisfy both of these inequalities. Only in these cases can problems I, II, II' have solutions.

Making this discussion, which presents no difficulty, we find that problems I, II, II' can have solutions only in the following 9 cases:

We will discuss these 9 cases further. This discussion is based on the actual determination of the numbers $\mathrm{M}^{\prime},\mathrm{M}^{\prime\prime}$ .
41. 1. - Case I. We have $v=s-1,s$ being a natural number. Based on formulas (116) we have
(121)

M_{1}^{\prime}=\max_{i=0,1,\ldots,\left[\frac{s-1}{2}\right]}\{G(2i)\},M_{1}^{\prime\prime}=\min_{i=1,2,\ldots,\left[\frac{s}{2}\right]}\{G(2i-1)\}

Taking into account (118) we find
$\mathrm{G}(2i)=i+1-\frac{2(i+s+2)}{2s+1}=\frac{i(s-1-i)+2s+1}{2s+1},i=0,1,\ldots,\left[\frac{s-1}{2}\right]$
$\left.\mathrm{G}(2i-1)=i-\frac{(2i-1)(2i+2s+3)}{4(2s+1)}=\frac{4i(s-i)+2s+3}{4(2s+1)},i=1,2,\ldots,\frac{s}{2}\right\rfloor$
A direct calculation then gives us

		$\displaystyle\mathbf{M}_{1}^{\prime}=\left\{\begin{array}[]{l}\frac{s^{2}+6s+5}{4(2s+1)},\quad\text{ pentru }s\text{ impar }\\ \frac{s^{2}+6s+4}{4(2s+1)},\end{array}\text{ pentru }s\right.\text{ par, }$
		$\displaystyle\mathbf{M}_{1}^{\prime\prime}=\frac{6s-1}{4(2s+1)}.$

Based on formulas (114) we have

M_{2}^{\prime}=\frac{(s+2)^{2}}{4(2s+1)},\quad M_{2}^{\prime\prime}=\frac{2s+3}{4(2s+1)}

$M^{\prime\prime}=M_{2}^{\prime\prime}$ It is easy to
see that $M_{1}^{\prime}>M_{2}^{\prime},\quad M_{1}^{\prime\prime}\geq M_{2}^{\prime\prime}$ It therefore follows $M^{\prime}=M_{1}^{\prime\prime}$ ,
It is worth noting that this result remains valid for $s=1$ although in this case the second formula (121) does not exist. However, if $s=1$ , we have $M^{\prime}=M_{1}^{\prime\prime}=G(0)=1$ .

Finally

M^{\prime}-M^{\prime\prime}=\left\{\begin{array}[]{l}\frac{s^{2}+4s+2}{4(2s+1)},\text{ dacă }s\text{ este impar, }\\ \frac{s^{2}+4s+1}{4(2s+1)},\text{ dacă }s\text{ este par. }\end{array}\right.

It is immediately verified that problems I, II, II' have solutions if and only if $s=1,2,3,4$ .
41. 2. - Case II. We have $v=3s-1$ . Proceeding as above we have
$\mathrm{M}_{1}^{\prime}=\max_{i=0,1,\ldots,s-1}\{\mathrm{G}(3i)\},\quad\mathrm{M}_{1}^{\prime\prime}=\min_{i=1,2,\ldots,s}\{\mathrm{G}(3i-1)\}$ where
$\mathrm{G}(3i)=i+1-\frac{i(3i+3s+5)}{2(3s+1)}=\frac{3i(s-1-i)+2(3s+1)}{2(3s+1)},i=0,1,\ldots,s-1$ , $\mathrm{G}(3i-1)=i-\frac{(3i-1)(3i+3s+4)}{6(3s+1)}=\frac{3i(3s-1-3i)+3s+4}{6(3s+1)},i=1,2,\ldots,s$ , from where

		$\displaystyle\mathbf{M}_{1}^{\prime}=$
		$\displaystyle\mathbf{M}_{1}^{\prime\prime}=\frac{2}{3(3s+1)}$

We also have

		$\displaystyle\mathbf{M}_{2}^{\prime}=$
		$\displaystyle\mathbf{M}_{2}^{\prime\prime}=\frac{3s+4}{6(3s+1)}$

It is easy to see that $M_{1}^{\prime}>M_{2}^{\prime},M_{2}^{\prime\prime}>M_{1}^{\prime\prime}$ so that $M^{\prime}=M_{1}^{\prime},M^{\prime\prime}=M_{1}^{\prime\prime}$ and we deduce

M^{\prime}-M^{\prime\prime}=\begin{cases}\frac{9s^{2}+54s+17}{24(3s+1)},&\text{ dacă }s\text{ este impar }\\ \frac{9s^{2}+54s+8}{24(3s+1)},&\text{ dacă }s\text{ este par }\end{cases}

Problems I, II, II' have solutions if and only if $s=1,2$ .
41. 3. - Case III. We have $v=3s$ and proceeding as in case II,
$\mathbf{M}_{\mathbf{1}}^{\prime}=\max\quad\{\mathrm{G}(3i)\},\mathbf{M}_{\mathbf{1}}^{\prime\prime}=\min\quad\{\mathrm{G}(3i-1)\}$
Spirit
$\square(3i)=i+1-\frac{3i(i+s+2)}{2(3s+2)}=\frac{i(3s-2-3i)+2(3s+2)}{2(3s+2)},\quad i=0,1,\ldots,s$ , $G(3i-1)=i-\frac{(3i-1)(3i+3s+5)}{6(3s+2)}=\frac{9i(s-i)+3s+5}{6(3s+2)},\quad i=1,2,\ldots,s_{n}$

We deduce from here

		$\displaystyle\mathbf{M}_{1}^{\prime}=\left\{\begin{array}[]{l}\frac{3s^{2}+20s+16}{8(3s+2)},\quad\text{ dacă }s\text{ este par },\\ \frac{3s^{2}+20s+17}{8(3s+2)},\quad\text{ dacă }s\text{ este impar, }\end{array}\right.$
		$\displaystyle\mathbf{M}_{1}^{\prime\prime}=\frac{3s+5}{6(3s+2)}.$

We also have

		$\displaystyle\mathbf{M}_{2}=\left\{\begin{array}[]{l}\frac{3(s+2)^{2}}{8(3s+2)},\quad\text{ dică }s\text{ este par, }\\ \frac{(3s+5)(3s+7)}{24(3s+2)},\quad\text{ dacă }s\text{ este impar, }\end{array}\right.$
		$\displaystyle\mathbf{M}_{2}=\frac{3s+5}{6(3s+2)}$

It is seen that $M_{1}^{\prime}>M_{2}^{\prime},\quad M_{1}^{\prime\prime}=M_{2}^{\prime\prime}$ , so $M^{\prime}=M_{1}^{\prime},M^{\prime\prime}=M_{1}$ and debuchern
$\mathbf{M}^{\prime}-\mathbf{M}^{\prime\prime}=\begin{cases}\frac{9s^{2}+48s+28}{24(3s+2)},&\text{ dacă }s\text{ este par, }\\ \frac{9s^{2}+48s+31}{24(3s+2)}&\text{ dacă }s\text{ este impar, }\end{cases}$
Problems I, II, II' have solutions if and only if $s=1,2,3$ .
41. 4.-Cases IV-IX. We can treat these 6 cases together.. Calculating on $\mathrm{M}_{1}^{\prime}\mathrm{M}_{1}^{\prime\prime}$ we find the following values

To show how this table is obtained, it will be sufficient to perform the calculations in one of the cases. We will choose case IX for this. We then have $v=41$ and $\mathrm{R}(n)=\frac{n(n+22)}{208}$ . For calculating numbers $M_{1}^{\prime},M_{1}^{\prime\prime}$ we use the formulas (116). Nonnegative integers of the form $8x+13y$ which checks inequalities $0\leqq 8x+13y\leqq 41$ are

0,8,13,16,21,24,26,29,32,34,37,39,40.

Based on Lemma 14 and formula (115), the values of $N(n)$ for these successive values of $n$ are the consecutive natural numbers from 1 to 13 inclusive. Based on the formula $\mathrm{G}(n)=\mathrm{N}(n)-\mathrm{R}(n)$ we obtain the values of $G(n)$ for his values $n$ which intervene in its determination $M_{1}^{\prime}$ We thus obtain the following

Table 1: Table 3.

$n$	0	8	13	16	21	24	26	29	32	34	37	39	40
$\mathrm{\penalty 10000\ N}(n)$	1	2	3	4	5	6	7	8	9	10	11	12	13
$\left.\mathrm{R}^{\prime}n\right)$	0	$\frac{15}{13}$	$\frac{35}{16}$	$\frac{38}{13}$	$\frac{903}{208}$	$\frac{69}{13}$	6	$\frac{1479}{208}$	$\frac{108}{13}$	$\frac{119}{13}$	$\frac{2183}{208}$	$\frac{183}{16}$	$\frac{155}{13}$
$\mathrm{G}(n)$	1	$\frac{11}{13}$	$\frac{13}{16}$	$\frac{14}{13}$	$\frac{137}{208}$	$\frac{9}{13}$	1	$\frac{185}{208}$	$\frac{9}{13}$	$\frac{11}{13}$	$\frac{105}{208}$	$\frac{9}{16}$	$\frac{14}{13}$

It follows from this that $\mathrm{M}_{1}=\frac{14}{13}$
To determine the $\mathrm{M}_{1}^{\prime\prime}$ we observe that non-negative integers of the form $8x+13y-1$ which checks inequalities $0\leqq 8x+13y-1\leqq 41$ are

7,12,15,20,23,25,28,31,33,36,38,39,41.

For the same reasons as above, the values of $N(n)$ for these successive values of $n$ are the consecutive natural numbers from 1 to 13 inclusive. We thus obtain

Table 2: Table 4.

$n$	7	12	15	20	23	25	28	31	33	36	38	39	41
$\mathrm{N}(\mathrm{n})$	1	2	3	4	5	6	7	8	9	10	11	12	13
$\mathrm{R}(n)$	203	51	555	105	1035	1175	175	1643	1815	261	285	183	2583
$\mathrm{R}(n)$	208	26	208	26	208	20”	26	208	208	26	26	16	208
$\mathrm{G}(n)$	$\frac{5}{208}$	$\frac{1}{26}$	$\frac{69}{288}$	$-\frac{1}{26}$	$\frac{5}{208}$	$\frac{73}{208}$	$\frac{7}{26}$	$\frac{21}{208}$	$\frac{57}{208}$	$-\frac{1}{26}$	$\frac{1}{26}$	$\frac{9}{16}$	$\frac{121}{208}$

It follows from this that $\mathbf{M}_{1}^{\prime\prime}=-\frac{1}{26}$
Returning to table 1, we see that in cases IV, V, VIII, IX we have . $M_{1}^{\prime}-M_{1}^{\prime\prime}>1$ and therefore all the more so $M^{\prime}-M^{\prime\prime}>1$ In these cases, problems I, II, II' have no solution.

Cases VI and VII still remain to be examined further. Formulas (114) give us

Table 3: Table 5.

case	$M_{2}^{\prime}$	$M_{2}^{\prime\prime}$
WE	$\frac{3}{5}$	$\frac{6}{35}$
VII	$\frac{49}{80}$	$\frac{13}{80}$

It follows, comparing with table 1, that
in case VI: $M^{\prime}=1,M^{\prime\prime}=\frac{1}{35},\quad M^{\prime}-M^{\prime\prime}=\frac{34}{35}<1$ ,
in case VII: $\mathrm{M}^{\prime}=1,\mathrm{M}^{\prime\prime}=\frac{1}{10},\quad\mathrm{M}^{\prime}-\mathrm{M}^{\prime\prime}=\frac{9}{10}<1$ .
In these cases, problems I, II, II' therefore have solutions.
In short, we have:
THEOREM 7. -In the case $m=3$ problems I, II, II' have solutions in and only in the following 18 cases, numbered from $1^{\circ}$ the $18^{\circ}$ .

Table 4: Table 6.

case	$1^{\circ}$	$2^{\circ}$	$3^{\circ}$	$4^{\circ}$	$5^{\circ}$	$6^{\circ}$	$7^{\circ}$	$8^{\circ}$	$9^{\circ}$
$b$ and $c$	1,1	1,2	1,3	1,4	1,5	1,6	1,7	2,3	2,5

case	$10^{\circ}$	$11^{\circ}$	$12^{\circ}$	$13^{\circ}$	$14^{\circ}$	$15^{\circ}$	$16^{\circ}$	$17^{\circ}$	$18^{\circ}$
$b$ and $c$	2,7	2,9	3,4	3,7	3,5	3,8	3,11	5,7	5,8

In all cases at least one of the coefficients of equation (1) is equal to 1, and the other two coefficients b and c are given in the accompanying table 6.
42. - We will also give the solutions to problems I, II, II' in the 18 cases $1^{\circ}-18^{\circ}$ highlighted.

The following table shows the values of $M^{\prime},M^{\prime\prime},M^{\prime}-M^{\prime\prime}$ and of the polynomial $\mathrm{R}(n)$ in the form of $\frac{n(n+b+c+1)}{2bc}$ Polynomials $\mathrm{P}(n),\mathrm{Q}(n),\mathrm{S}(n)$ can be obtained in the form $R(n)+\frac{\lambda}{2bc}$ and in the table the range of variation is given $\lambda$ for the three polynomials.

Table 5: Table 7.

case	$\mathrm{M}^{\prime}$	$M^{\prime\prime}$	$M^{\prime}-M^{\prime\prime}$	R (n)	summer interview of $\lambda$ for
case	$\mathrm{M}^{\prime}$	$M^{\prime\prime}$	$M^{\prime}-M^{\prime\prime}$	R (n)	$\mathrm{P}(n)$	Q(n)	S (n)
$1^{\circ}$	1	1	0	$\frac{n(n+3)}{2}$	$(1,3)$	[2,4)	$(0,2]$
$2^{\circ}$	1	$\frac{3}{4}$	$\frac{1}{4}$	$\frac{n(n+4)}{4}$	$(2,5)$	$[4,7)$	(0,3]
$3^{\circ}$	1	$\frac{2}{3}$	$\frac{1}{3}$	$\frac{n(n+5)}{6}$	$(3,7)$	$[6,10)$	$(0,4]$
$4^{\circ}$	$\frac{9}{8}$	$\frac{5}{8}$	$\frac{1}{2}$	$\frac{"(n+6)}{\gamma}$	$(5,9)$	$[9,13)$	$(1,5]$
$5^{\circ}$	$\frac{6}{5}$	$\frac{3}{0}$	$\frac{3}{5}$	$\frac{n(n+7)}{10}$	$(7,11)$	$[12,16)$	(2,6]
$6^{\circ}$	$\frac{4}{3}$	$\frac{7}{12}$	$\frac{3}{4}$	$\frac{n(n+8)}{12}$	$(10,13)$	[16,19 ;	$(4,7]$
$7^{\circ}$	$\frac{10}{7}$	$\frac{4}{7}$	$\frac{6}{7}$	$\frac{n(n+9)}{1+}$	$(13,15)$	$[20,22)$	(6,8]
$8^{\circ}$	1	$\frac{5}{12}$	$\frac{7}{12}$	$\frac{n(n+6)}{12}$	$(6,11)$	$[12,17)$	$(0,5]$
$9^{\circ}$	1	$\frac{7}{20}$	$\frac{13}{20}$	$\frac{n(n+3)}{20}$	$(10,17)$	$[20,27)$	(0,7]
$10^{\circ}$	$\frac{8}{1}$	$\frac{9}{28}$	$\frac{25}{28}$	$\frac{n(n+10)}{28}$	$(18,23)$	$[32,37)$	$(4,9]$
$11^{\circ}$	$\frac{10}{9}$	$\frac{11}{36}$	$\frac{29}{36}$	$\frac{n(n+12)}{36}$	$(22,29)$	$[40,47)$	(4,11]
$12^{\circ}$	1	$\frac{1}{6}$	$\frac{5}{6}$	$\frac{n(n+8)}{24}$	$(12,16)$	$[24,28)$	(0,4]
$13^{\circ}$	1	$\frac{4}{21}$	$\frac{19}{21}$	$\frac{n(n+11)}{42}$	$(21,25)$	[42,46	(0,4
$14^{\circ}$	1	4 10	$\frac{11}{10}$	$\frac{n(n+9}{30}$	$(15,23)$	$[30,38)$	(0,8

case	$M^{\prime}$	M"	$\mathrm{M}^{\prime}-\mathrm{M}^{\prime\prime}$	R (n)	interv. . is his/her var. $\lambda$ for
case	$M^{\prime}$	M"	$\mathrm{M}^{\prime}-\mathrm{M}^{\prime\prime}$	R (n)	$\mathrm{P}(n)$	Q(n)	S (n)
$15^{\circ}$	$\frac{17}{16}$	$\frac{11}{4\times}$	$\frac{5}{6}$	$\frac{n(n+12)}{48}$	(27,35)	$[51,59)$	(3,11]
$16^{\circ}$	$\frac{13}{11}$	$\frac{7}{33}$	$\frac{32}{33}$	$\frac{n(n+15)}{66}$	$(45,47)$	[(8,80)	(12,14]
$17^{\circ}$	1	$\frac{1}{35}$	$\frac{34}{35}$	$\frac{n(n+13)}{70}$	$(35,37)$	$[70,72)$	(0,2]
$18^{\circ}$	1	$\frac{1}{10}$	$\frac{9}{10}$	$\frac{n(n+14)}{80}$	$(40,48)$	$[80,88)$	$(0,8]$

Based on the forms (69) the interval relative to the polynomial $Q(n)$ is deduced from the interval relative to the polynomial $\mathrm{P}(n)$ by a translation equal to $bc$ and the interval relative to the polynomial S(n) by a translation equal to - bc. The interval relative to $Q(n)$ is closed on the left, and the relative one on $S(n)$ is closed on the right, the respective extremities corresponding to the exceptional solutions defined in No. 25.

The exceptional solutions of problems II, II' are given in the following table.

Table 6: Table 8.

case

In Q(n)

ener

S (n)

1^{\circ}

(n+1)(n+2)

(n+1)(n+2)

2^{\circ}

(n+2)^{2}

(n+1)(n+3)

3^{\circ}

\frac{(n+2)(n+3)}{6}

(n+1)(n+4)

4^{\circ}

\frac{(n+3)^{2}}{8}

(n+1)(n+5)

5^{\circ}

(n+3)(n+4)

(n+1)(n+6)

6^{\circ}

\frac{(n+4)^{2}}{12}

(n+1)(n+7)

7^{\circ}

\frac{(n+4)(n+5)}{14}

(n+1)(n+8)

8^{\circ}

\frac{n^{2}+6n+12}{12}

(n+1)(n+5)

9^{\circ}

n^{2}+8n+20

(n+1)(n+7)

43.
- —
  
  Interesting particular solutions are those in which e.g. the polynomial $\mathrm{P}(n),\mathrm{Q}(n)$ or $\mathrm{S}(n)$ decomposes into the product of two linear factors with rational coefficients. From Table 7 it can be deduced that there are an infinity of such polynomials $\mathrm{P}(n)$ and an infinity of such polynomials $\mathrm{S}(n)$ in all 18 cases. There are an infinity of such polynomials Q(n) in the cases $1^{\circ},3^{\circ},5^{\circ},7^{\circ}$ , there is a sure one (which is obviously the exceptional solution) in the cases $2^{\circ},4^{\circ},6^{\circ}$ and there is none in the other 11 cases.

To prove these statements, it is sufficient to first observe that the polynomial $n(n+b+c+1)+\lambda$ can be decomposed into two first-degree factors with rational coefficients only if
$(b+c+1)^{2}-4\lambda\geqq 0$ We now find, on the
one hand, that we have an infinity, one or no polynomial $\mathrm{P}(n),\mathrm{Q}(n)$ or $\mathrm{S}(n)$ of the sought-after form as the range of variation of $\lambda$ corresponding to the interval
$\left(-\infty,\frac{(b+c+1)^{2}}{4}\right]o$ infinity, one or no common point. On the other hand, the values of $\frac{(b+c+1)^{2}}{4}$ for the 18 cases are successively equal to $\frac{9}{4},4,\frac{25}{4},9,\frac{49}{4},16,\frac{81}{4},9,16,25,36,16,\frac{121}{4},\frac{81}{4},36,\frac{225}{4}$ , $\frac{169}{4},49$ .
44. - To highlight some of these solutions we will examine three particular forms.
I. Let's look for the polynomial $\mathrm{P}(n),\mathrm{Q}(n)$ or $\mathrm{S}(n)$ where the difference of the roots is an integer. The polynomial sought is then of the form $\frac{(n+u)(n+u+v)}{2bc}$ , where $u$ is a rational number, and $v$ an integer that can be assumed to be non-negative.

It follows from this that we must have
somewhere to start.

2u+v=b+c+1,\quad u(u+v)=\lambda

(b+c+1)^{2}-v^{2}=4\lambda

We immediately see that we have a finite number of solutions, $\lambda$ ₇ must appear within the range given in table 7.

By doing all the calculations, we find the solutions contained in the following table.

Table 7: Table 9.

case

P (n)

Q (n)

S (n)

1^{\circ}

\begin{aligned} &\frac{(2n+3)^{2}}{8},\frac{(n+1)(n+2)}{2},\\ &\frac{(2n+1)(2n+5)}{8}\end{aligned}

\begin{aligned} &\frac{(2n+3)^{2}}{8},\\ &\frac{(n+1)(n+2)}{2}-\end{aligned}

\begin{aligned} &\frac{(n+1)(n+2)}{2}\\ &\frac{(2n+1)(2n+5)}{8}\end{aligned}

2^{\circ}

\begin{aligned} &\frac{\left(n+2^{2}\right)}{4},\frac{(2n+3)(2n+5)}{10},\\ &\frac{(n+1)(n+3)}{4}\end{aligned}

\frac{(n+2)^{2}}{4}

\begin{aligned} &\frac{(n+1)(n+3)}{4}\\ &\frac{(2n+1)(2n+7)}{16}\end{aligned}

3^{\circ}

\begin{aligned} &\frac{(2n+5)^{2}}{24},\frac{(n+2)(n+3)}{6},\\ &\frac{(2n+3)(2n+7)}{24},\frac{(n+1)(n+4)}{6}\end{aligned}

\begin{aligned} &\frac{(2n+5)^{2}}{24},\\ &\frac{(n+2)(n+3)}{6}\end{aligned}

\begin{aligned} &\frac{(n+1)(n+4)}{6},\\ &\frac{(2n+1)(2n+9)}{24}\end{aligned}

4^{\circ}

\begin{aligned} &\frac{(2n+5)(2n+7)}{32},\frac{(n+2)(n+4)}{8},\\ &\frac{(2n+3)(2n+9)}{32}\end{aligned}

\frac{(n+3)^{2}}{8}

\begin{aligned} &\frac{(n+1)(n+5)}{8},\\ &2n+1)(2n+11)\end{aligned}

5^{\circ}

\begin{gathered}\frac{(n+2)(n+5)}{10},\\ \frac{(2n+3)(2n+11)}{40}\end{gathered}

\begin{aligned} &\frac{(2n+7)^{2}}{40},\\ &\frac{(n+3)(n+4)}{10}\end{aligned}

\frac{(n+1)(n+6)}{10}

(2n+1)(2n+13)

6^{\circ}

\frac{(n+2)(n+6)}{12}

\frac{(n+4)^{2}}{12}

\frac{(n+1)(n+7)}{12}

7^{\circ}

\frac{(n+2)(n+7)}{14}

\begin{aligned} &\frac{(2n+9)^{2}}{56}\\ &\frac{(n+4)(n+5)}{14}\end{aligned}

\frac{(n+1)(n+8)}{14}

8^{\circ}

\begin{aligned} &\frac{(n+3)^{2}}{12},\frac{(2n+5)(2n+7)}{48}\\ &\frac{(n+2)(n+4)}{12},\frac{(2n+3)(2n+9)}{48}\end{aligned}

\begin{aligned} &\frac{(n+1)(n+5)}{12}\\ &\frac{(2n+1)(2n+11)}{48}\end{aligned}

9^{\circ}

\begin{aligned} &\frac{(n+4)^{2}}{20},\frac{(2n+7)(2n+9)}{80},\\ &\frac{(n+3)(n+5)}{20},\frac{(2n+5)(2n+11)}{80},\\ &\frac{(n+2)(n+6)}{20}-\end{aligned}

\begin{aligned} &\frac{(n+1)(n+7)}{20}\\ &\frac{(2n+1)(2n+15)}{80}\end{aligned}

Table 8: Table 9 (continued)

case	P (n)	Q(n)	S (n)
$10^{\circ}$	$\begin{aligned} &\frac{(2n+7)(2n+13)}{112},\frac{(n+3)(n+7)}{28},\\ &\frac{(2n+5)(2n+15)}{112}\end{aligned}$	-	$\begin{aligned} &\frac{(n+1)(n+9)}{28}\\ &\frac{(2n+1)(2n+19)}{112}\end{aligned}$
$11^{\circ}$	$\begin{aligned} &\frac{(n+3)(n+9)}{36},\\ &\frac{(2n+5)(2n+19)}{144}\end{aligned}$	-	$\begin{gathered}\frac{(n+1)(n+11)}{36},\\ \frac{(2n+1)(2n+23)}{144}\end{gathered}$
$12^{\circ}$	$\begin{aligned} &\frac{(2n+7)(2n+9)}{96},\frac{(n+3)(n+5)}{24}\\ &\frac{(2n+5)(2n+11)}{96}\end{aligned}$	-	$\frac{(2n+1)(2n+15)}{96}$
$13^{\circ}$	$\frac{(n+3)(n+8)}{42},\frac{(2n+5)(2n+17)}{168}$	-	-
$14^{\circ}$	$\begin{aligned} &\frac{(2n+9)^{2}}{120},\frac{(n+4)(n+5)}{30},\\ &\frac{(2n+7)(2n+11)}{120},\frac{(n+3)(n+6)}{30},\\ &\frac{(2n+5)(2n+13)}{120}\end{aligned}$	-	$\begin{aligned} &\frac{(n+1)(n+8)}{30},\\ &\frac{(2n+1)(2n+17)}{120}\end{aligned}$
$15^{\circ}$	$\begin{aligned} &\frac{(2n+9)(2n+15)}{192},\frac{(n+4)(n+8)}{4\gamma}\\ &\frac{(2n+7)(2n+17)}{192}\end{aligned}$	-	$\begin{aligned} &\frac{(n+1)(n+11)}{48}\\ &\frac{(2n+1)(2n+23)}{192}\end{aligned}$
$16^{\circ}$	-	-	$\frac{(n+1)(n+14)}{66}$
$17^{\circ}$	$\frac{(n+4)(n+9)}{70}$	-	-
$18^{\circ}$	$\begin{aligned} &\frac{(2n+11)(2n+17)}{320},\frac{(n+5)(n+9)}{80},\\ &\frac{(2n+9)(2n+19)}{320}\end{aligned}$	-	$\frac{(2n+1)(2n+27)}{32v}$

This table contains in particular all the solutions in which the polynomials $\mathrm{P}(n),\mathrm{Q}(n),\mathrm{S}(n)$ are perfect squares.
II. Let's look for the solutions in which the polynomial $\mathrm{P}(n),\mathrm{Q}(n)$ or $\mathrm{S}(n)$ is, apart from an (obviously) rational factor, a product of two consecutive integers, whatever they may be $n$ .

The polynomial sought is then of the form $(un+v)(un+v+1)$ , where $u$ is a natural number, $v$ an integer and $w$ a rational number. By identification we find

\frac{u^{2}}{w}=\frac{1}{2bc},\quad\frac{u(2v+1)}{w}=\frac{b+c+1}{2bc},\quad\frac{v(v+1)}{w}=\frac{\lambda}{2bc}.

From here it follows $w=2bcu^{2}$ , so $w$ is a natural number. We still have

u(b+c+1)=2v+1,\quad\lambda=\frac{(b+c+1)^{2}u^{2}-1}{4u^{2}}

(122)

From the first of these formulas it follows that $u$ and $b+c+1$ must be odd numbers. It follows then that $v$ is a natural number. It also follows that we can only have solutions in the cases $1^{\circ},3^{\circ},5^{\circ},7^{\circ},13^{\circ},14^{\circ},16^{\circ},17^{\circ}$ .

The variation interval of the second member of the second formula (122) is

\left(\frac{(b+c)(b+c+2)}{4},\frac{(b+c+1)^{2}}{4}\right).

(123)

We will have as many solutions as the values of the second member of the second formula (112) fall within the variation interval of $\lambda$ for that polynomial $\mathrm{P}(n),Q(n),\mathrm{S}(n)$ .
III. Let us also look for solutions in which the polynomial $\mathrm{P}(n),\mathrm{Q}(n)$ or $\mathrm{S}(n)$ is, apart from an (obviously) rational factor, the product of two consecutive integers of the same parity.

The polynomial sought is then of the form $(un+v)(un+v+2)$ where $u$ is a natural number, $v$ an integer and $w$ a rational number. We now have

\frac{u^{2}}{w}=\frac{1}{2bc},\frac{2u(v+1)}{w}=\frac{b+c+1}{2bc},\quad\frac{v(v+2)}{w}=\frac{\lambda}{2bc}

and it is deduced again $w=2bcu^{2}$ as well as

u(b+c+1)=2(v+1),\lambda=\frac{(b+c+1)^{2}u^{2}-4}{4u^{2}}

(124)

It is seen again that $v$ is a natural number. However, we must distinguish two cases. If $b+c+1$ is odd, so in the cases $1^{\circ},3^{\circ},5^{\circ},7^{\circ},13^{\circ}$ , $14^{\circ},16^{\circ},17^{\circ},u$ must be even ( $\geq 2$ ) and the range of variation of the second member of the second formula (124) is still (123). However, if $b+c+1$ is even, so in the other 10 cases, $u$ can be
any mature number and the range of variation of the second member of the second formula (124) is
(125)

\left(\frac{(b+c-1)(b+c+3)}{4},\quad\frac{(b+c+1)^{2}}{4}\right)

The effective solutions are determined as in case II.
To find the solutions in cases II. III, first form the table of intervals (123), (125).

Table 9: Table 10.

case	$1^{\circ}$	$2^{\circ}$	$3^{\circ}$	$4^{\circ}$	$5^{\circ}$	$6^{\circ}$
interval (123)	$\left[2,\frac{9}{4}\right)$		$\left(6,\frac{25}{4}\right)$		$\left.\mid 12,\frac{49}{4}\right)$
interval (125)		$[3,4)$		$[8,9)$		$[15,16)$

case	$7^{\circ}$	$8^{\circ}$	$9^{\circ}$	$10^{\circ}$	$11^{\circ}$	$12^{\circ}$
interval (123)	$\left[\angle 0,\frac{81}{4}\right)$
interval (125)		$[8,9)$	$[15,16)$	$[24,25)$	$[35,36)$	$[15,16)$

case	$13^{\circ}$	$14^{\circ}$	$15^{\circ}$	$16^{\circ}$	$11^{\circ}$	$18^{\circ}$
interval (123)	$\left(30,\frac{121}{4}\right)$	$\left.20,\frac{81}{4}\right)$		$\left.56,\frac{225}{4}\right)$	$\left.42,\frac{169}{4}\right)$
interval (125)			$[35,3)$			$[48,49)$

From examining this table and taking into account the range of variation of $\lambda$ given in table 7, it follows that in case II there are an infinity of polynomials $\mathrm{P}(n)$ of this kind in cases $1^{\circ},3^{\circ}$ and $14^{\circ}$ an infinity of polynomials $Q(n)$ of this kind in cases $1^{\circ},3^{\circ},5^{\circ},7^{\circ}$ and a single polynomial $\mathrm{S}(n)$ similarly in the case $1^{\circ}$ .

In case III there are an infinity of polynomials $\mathrm{P}(n)$ of this kind in cases $1^{\circ},2^{\circ},3^{\circ},4^{\circ},8^{\circ},9^{\circ},12^{\circ},14^{\circ},0$ infinity of polynomials $Q(n)$ of this kind in cases $1^{\circ},3^{\circ},5^{\circ},7^{\circ}$ and a single polynomial $S(n)$ thus in cases $1^{\circ},2^{\circ}$ . Not all of these solutions are different from those found in case II. Namely those in which $u$ and $v$ are both even, they obviously reduce,
by simplification by 4, to the solutions found in case II. Conversely, amplicisantos) by 4, from any solution of case II a solution of case III is deduced.

The distinct solutions in cases II, III are included in the following table

Table 10: Table 11.

case	P(n)	Q (n)	S (n)
$1^{\circ}$	$\frac{(2n+5)^{2}t^{2}-1}{8t^{2}}$	$\frac{(2n+3)^{2}t^{2}-1}{8t^{2}}$	$\frac{(n+1)(n+2)}{2}$
$2^{\circ}$	$\frac{(n+2)^{2}t^{2}-1}{4t^{2}}$	-	$\frac{(n+1)(n+3)}{4}$
$3^{\circ}$	$\frac{(2n+5)^{2}t^{2}-1}{24t^{2}}$	$\frac{(2n+5)^{2}t^{2}-1}{24t^{2}}$	-
$4^{\circ}$	$\frac{(n+3)^{2}t^{2}-1}{8t^{2}}$	-	-
$5^{\circ}$	-	$\frac{(2n+7)^{2}t^{2}-1}{40t^{2}}$	-
$7^{\circ}$	-	$\frac{(2n+9)^{2}t^{2}-1}{56t^{2}}$	-
$8^{\circ}$	$\frac{(n+3)^{2}t^{2}-1}{12t^{2}}$	-	-
$9^{\circ}$	$\frac{(n+4)^{2}t^{2}-1}{20t^{2}}$	-	-
$12^{\circ}$	$\frac{(n+4)^{2}t^{2}-1}{24t^{2}}$	-	-
$14^{\circ}$	$\frac{(2n+9)^{2}t^{2}-1}{120t^{2}}$	-	-

where $t$ is an arbitrary natural number.
Various particular properties can be stated on which it is useless to insist.

Faculty of Mathematics and Physics of the "V. Babes" University of Chuj

BIBLIOGRAPHY

1.

Euler L., Introduction to the Analysis of Infinites or Opera Omnia, S. 1, T. 8.

Sylvester J. J., Outlines of seven leotures on the partitions of numbers. Proc. Londor Math. Soc., 28, 33-96 (1897).
2b. Sylvester J. ², discovery in the partition of mumbers , Quarterly J. of Math., 3a W. Wrol

Coefficients, Zeitsch number of solutions of dophantic equations with coprime coefficients, Zeîtsch, für Math. u Ph., 20, 97–111(1875).

3b. Weihrauch K., On the expressions $\Sigma$ fn (m) and the transformations of the formula for solution numbers, … ibid., 20, 111-117 (1875)
4. Skolem Th., Nota 14 la cartea Lehrbuch der Combinatorik de Dr. E. Netto, ed. II. 1927.
5. Gelfond AO, Iscislevie konecinîh ramostei, Moskva-Leningrad, 1952.
6. Dedekind R., On the decomposition of numbers by their greatest common divisors. Works, II, 103-147, (1931).

On a problem of number partitioning

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ON A PROBLEM OF PARTITION OF NUMBERS

§ 1.

So finally we have

§ 2.

§ 3.

§ 5.

24. - We can now demonstrate

§ 6.

§7.

§ 8.

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