1. 1.

   An arithmetic function$f(n)$Therefore, a (real) function defined on the set of natural numbers is said to be multiplicative if we have
   (1)

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regardless of the (natural) numbers$a,b$first among themselves.
In the following we will designate by ($a_{1}, a_{2}, ..., a_{n}$) the gcd and by$\left[a_{1},a_{2},\ldots,a_{n}\right]$the least common multiple (LCM) of (natural) numbers$a_{1}, a_{2}, ..., a_{n}$The numbers$a_{1}, a_{2}, ..., a_{n}$are relatively prime if and only if$\left(a_{1},a_{2},\ldots,a_{n}\right)=1$With these notations, the multiplicative property of the function$f(n)$is written$(a,b)=1\Rightarrow f(ab)=f(a)f(b)$2.
In the well-known book by G. PÓlya and G. SZEGÓ [3, p. 126] an interesting property of multiplicative arithmetic functions is stated and proven. In a slightly modified form, we can state this property as follows:

THEOREM, 1. If$f(n)$is a multiplicative arithmetic function and$a_{1},a_{2},\ldots,a_{n}$For any natural numbers, we have formulas
(2)$\quad f\left(\left[a_{1},a_{2},\ldots,a_{n}\right]\right)=\prod_{k=1}^{n}\left(\prod_{(k)}f\left(\left(a_{i_{1}},a_{i_{2}},\ldots,a_{i_{k}}\right)\right)\right)^{(-1)^{k-1}}$
(3)

Or$\Pi$refers to a product extended to all$\binom{n}{k}$combinations$k$has$k,i_{1},i_{2}$..,$i_{k}$clues$1,2,\ldots,n$.
For$n=2$the property reverts to the formula

Or$f(n)$is a multiplicative arithmetic function and$a,b$are any two natural numbers. This property is also given as an exercise in my higher algebra course (lithographed) [5, p. 190].

The proof of Theorem 1 in [3] and also only of the special case$n=2$The proof in my cited course [5] is based on the well-known, unique prime factorization of a natural number. In what follows, we will provide a proof that does not rely on this factorization.
3. To begin, we will demonstrate the property in the particular case$n=2$, therefore we will establish formula (4).

We will first demonstrate
Lemma 1.$a,b$being two relatively prime numbers and$d$a third natural number, there exists a decomposition$d=d^{\prime}d^{\prime\prime}$of$d$in the product of two complementary divisors$d^{\prime},d^{\prime\prime}$such as one might have

We only consider positive divisors of numbers.
Let's move on to the proof of Lemma 1. There exist divisors of$d$who are first with$a$For example, the number 1 is such a divisor. So then$d^{\prime}$the largest of the divisors of$d$who is first with$a$and which exists since all the divisors of$d$, so in particular those that are prime with$a$, are$\leqq d$So then$d^{\prime\prime}$the complementary divisor of$d^{\prime}$It suffices to demonstrate the first and third formulas (5), the second being true by construction.

Either$e=\left(d^{\prime},d^{\prime\prime}\right)$and suppose that$e>1$. We have$\geqslant\mid d^{\prime\prime}$, hence$ed^{\prime}\mid d$Next$e\mid d^{\prime}$And$\left(d^{\prime},a\right)=1$it follows that$\left(ed^{\prime},a\right)=1$Finally, we have$d^{\prime}<ed^{\prime}$which contradicts the definition of$d^{\prime}$It follows that$e=1$, SO$\left(d^{\prime},d^{\prime\prime}\right)=1$.

Either$f=\left(d^{\prime\prime},b\right)$and suppose that$f>1$We then have$f\mid b$, hence$(f,a)=1$Next$f\mid d^{\prime\prime}$And$\left(d^{\prime},a\right)=1$it follows that$fd^{\prime}\mid d$therefore also that$\left(fd^{\prime},a\right)=1$We finally have$d^{\prime}<fd^{\prime}$which is still in contradiction with the definition of$d^{\prime}$It follows that$f=1$, SO$\left(d^{\prime\prime},b\right)=1$.

Lemma 1 is therefore proven.
Note 1. Every divisor$d_{1}$of$d$and who is first with$a$is a divisor of the greatest divisor$d^{\prime}$of$d$who enjoys the same property. Indeed
, of$d_{1}\left|d,d^{\prime}\right|d,\left(d_{1},a\right)=\left(d^{\prime},a\right)=1$it follows that$\left[d_{1},d^{\prime}\right]\mid d$And$\left(\left[d_{1},d^{\prime}\right],a\right)==\left[\left(d_{1},a\right),\left(d^{\prime},a\right)\right]=1$, done$\left(\left[d_{1},d^{\prime}\right],a\right)=1$. But$d^{\prime}\leqq\left[d_{1},d^{\prime}\right]\leqq d^{\prime}$, according to the definition of$d^{\prime}$It follows that$\left[d_{1},d^{\prime}\right]=d^{\prime}$, SO$d_{1}\mid d^{\prime}$.

Note 2. The divisor of has also been used in other problems by W. Sierpinski [6, p. 17].
4. Let us proceed to the proof of formula (4). Let$d=(a,b)$And$a=da^{\prime},b=db^{\prime}$. SO$\left(a^{\prime},b^{\prime}\right)=1$Let's apply Lemma 1 to numbers.$a^{\prime},b^{\prime},d$. We have$a=d^{\prime}d^{\prime\prime}a^{\prime},\quad b=d^{\prime\prime}d^{\prime}b^{\prime},\quad(a,b)=d^{\prime}d^{\prime\prime},\quad[a,b]==d^{\prime}b^{\prime}d^{\prime\prime}a^{\prime}$.

But, from$\left(d^{\prime},d^{\prime\prime}\right)=\left(d^{\prime},a^{\prime}\right)=1$it follows that$\left(d^{\prime},d^{\prime\prime}a^{\prime}\right)=1$Similarly, we have$\left(d^{\prime\prime},d^{\prime}b^{\prime}\right)=1$From property (1) of multiplicativity, it follows that
(6)

Similarly$\left(d^{\prime},d^{\prime\prime}a^{\prime}\right)=\left(b^{\prime},d^{\prime\prime}a^{\prime}\right)=1$it follows that$\left(d^{\prime\prime}a^{\prime},d^{\prime}b^{\prime}\right)=1$and we deduce from this

By comparing formulas (6) and (7) we obtain the desired formula (4).
5. Let's prove Theorem 1. To prove formula (2) we will proceed by complete induction on the number$n$numbers$a_{i}$. For$n=2$The formula reduces to (4) and is therefore proven. Suppose the property is true for$n-1$numbers and let's demonstrate it for$n$Numbers$a_{1},a_{2},\ldots,a_{n}(n>2)$To simplify the notation, let's...$E_{k}==\prod_{(k)}f\left(\left(a_{i_{1}},a_{i_{2}},\ldots,a_{i_{k}}\right)\right),k=1,2,\ldots,n$and let us designate by$E_{k}^{\prime},k=1,2$..
,$n-1$numbers$E_{k}$corresponding to$n-1$Numbers$a_{1},a_{2},\ldots,a_{n-1}$and by$E_{k}^{*},k=1,2,\ldots,n-1$the same numbers corresponding to$n-1$Numbers$\left(a_{1},a_{n}\right),\left(a_{2},a_{n}\right),\ldots,\left(a_{n-1},a_{n}\right)$Taking into account (4) we have

and it's easy to verify that

by asking$E_{n}^{\prime}=1,E_{0}^{*}=f\left(a_{n}\right)$As a result of the distributive property of operations$(a,b)$And$[a,b]$, We have

7 - Mathematica Vol. 13 (36) – Fasc. 2/1971

By hypothesis we have

and by virtue of equality (10) we also have

Considering (8), (9), (11), and (12), we deduce formula (2). Formula
(3) is proven in the same way.
Theorem 1 is therefore proven.
The arithmetic function$f(n):=n$is indeed multiplicative and formula (4) becomes$(a,b)[a,b]=ab$which can be demonstrated directly using the divisibility relation. The general formulas (2), (3) can easily be written in the case of the function$f(n)=n$. For formula (2) this is done in the cited book by g. pólya and g. szigGó [3].

1. 6.

   Euler's arithmetic function$\psi(n)$(the indicator) is indeed multiplicative and we have

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if (and only if)$(a,b)=1$
We know that $\varphi(n)$is equal to the number of natural numbers$\leq n$and first with$n$.

If the condition$(a,b)=1$If formula (13) is not verified, it is no longer true. In general, if$d=(a,b)$we have the formula

This formula, although not explicitly stated, can be found in various treatises. It suffices to cite E. Lucas [2, p. 398] and Le Dickson [1, p. 13].

The proofs of formula (14) have always been based on the prime factorization of numbers. There is still an advantage to dispensing with this factorization. This is what we will do in what follows.
7. We will first prove the

Lemma 2. If a|b we have
( 15 )

We will refer to$a\mid b$the fact that the number$b$is divisible by the number$a$, a notation that we have already used in the first part of this work.

We can give a proof of this lemma, analogous to the well-known proof of formula (13). This proof amounts to counting the terms of the sequence of$ab$prime natural numbers that are coprime with$ab$The key terms of this sequence are arranged, for clarity, in a rectangular table.

of$a$lines and$b$columns so that the element of the$(i+1)^{-\text{eme ligne }}$and the$j$-a column$ib+j$This demonstration can be found in all treatises on number theory. It suffices to cite the excellent book by$W$. CLERPINSKI [6, p. 229]. You can also consult my course already cited [5, p. 182].

To prove Lemma 2, let us consider again Table (16). We will show that, under the conditions of the lemma, we have

In short, let's note that we generally have$(\alpha,\beta\gamma)=1\Leftrightarrow(\alpha,\beta)=$:$=:(\alpha,\gamma)=1$And$(\alpha,\beta)=(\alpha,\beta+\gamma\alpha)$, when$\alpha,\beta,\gamma$are natural numbers. If now$(ib+j,ab)=1$It follows that$(ib+j,b)=1$, done also$(j,b)=1$We have done

Let us now suppose that$(j,b)=1$It follows that$(ib+j,b)=1$SO$(ib+j,a)=1$since a is a divisor of$b$It follows that$(ib+j,ab)=1$So we have

From (18), (19) we get formula (17). From formula (17) we get that in each row of table (16) there is exactly$\varphi(b)$prime numbers with$ab$Lemma 2 is therefore proven.

Moreover, the property expressed by Lemma 2 is equivalent to that expressed by the

Iremme 3. We have
(20)

To show that 1st 1st 3 results from 1st 2 i 1, it suffices to apply formula (15) to the numbers$a$,$ab$The second being clearly divisible by the first. To show that Lemma 2 follows from Lemma 3, either$a\mid b$We then have$b=ab^{\prime}$and by virtue of Lemma 3 we have$\varphi(ab)==\varphi\left(a^{2}b^{\prime}\right)=a\varphi\left(ab^{\prime}\right)=a\varphi(b)$This results in formula (15).

An immediate consequence of (20) is formula
(21)

whatever the natural numbers$a,b,m$8. Let
us demonstrate formula (14). By setting$d=(a,b)$, We have$d[a,b]=ab$. But$d[[a,b]$, SO$\varphi(ab)=d\varphi([a,b])$Taking into account formula (4), it follows that$\varphi(d)\varphi(ab)=d\varphi[(a,b])\varphi(d)=d\varphi(a)\varphi(b)$, hence formula (14).
9. Formula (14) can be generalized and we can state the

THEOREM 2.$a_{1},a_{2},\ldots,a_{n}$being natural numbers and setting$d=\left(a_{1},a_{2},\ldots,a_{n}\right)$, We have

Or$\Pi$has the same meaning as in Theorem 1.
(k)

We will perform a complete inductive demonstration on the number$n$numbers$a_{i}$. For$n=2$Formula (22) reduces to (14), which has already been proven. Now suppose that the property is true for$n-1$numbers and let's demonstrate - 1a for$n$numbers ($n>2$).