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Tiberiu Popoviciu

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T. Popoviciu

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T. Popoviciu, Sur les indicateurs, Bull. Sc. de l’Ecole Polytechnique de Timişoara, 3 (1930) nos. 1-2, pp. 72-80 (in French).

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Bull. Sc. de l’Ecole Polytechnique de Timişoara

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[JFM 56.0873.01]

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1930 d -Popoviciu- Bull. Sci. Ec. Polyt. Timisoara - On the indicators
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  1. The inverse problem of calculating the indicator consists in solving the equation:
    (1)
φ ( N ) = HAS . φ ( N ) = HAS . varphi(N)=A.\varphi(\mathrm{N})=\mathrm{A} .φ(N)=HAS.
First, note that it suffices to search for even numbers N satisfying (1). Indeed, from property 30, it follows that:
φ ( N ) = φ ( 2 N ) φ ( N ) = φ ( 2 N ) varphi(N)=varphi(2N)\varphi(\mathrm{N})=\varphi(2 \mathrm{~N})φ(N)=φ(2 N)
if N N NNNis odd. We will constantly use this remark in what follows.
But before addressing the solution of equation (1), we must examine whether it is possible or not. It is obvious that if A is odd and not equal to 1, the equation is impossible, but it is easy to see that it can be impossible even if A is even. For example, the relation:
φ ( N ) = 14 φ ( N ) = 14 varphi(N)=14φ(N) = 14φ(N)=14
is impossible.
More generally, one could propose solving the equation:
(2) φ i ( N ) = HAS (2) φ i ( N ) = HAS {:(2)varphi_(i)(N)=A:}\begin{equation*} \varphi_{i}(\mathrm{~N})=\mathrm{A} \tag{2} \end{equation*}(2)φi( N)=HAS
Given A. It is evident that if equation (2) is possible for i = m i = m i=mi=mi=m, all equations (2), where i = 1 , 2 , … m − 1 i = 1 , 2 , … m − 1 i=1,2,dots m-1i=1,2, \ldots m-1i=1,2,…m−1will be possible and if the equation with i = m i = m i=mi=mi=mis impossible, all equations with i = m + 1 i = m + 1 i=m+1i=m+1i=m+1, m + 2 , … m + 2 , … m+2,dotsm+2, \ldotsm+2,…will be impossible.
Yes, the equation:
φ i ( N ) = HAS φ i ( N ) = HAS varphi_(i)(N)=A\varphi_{i}(\mathrm{~N})=\mathrm{A}φi( N)=HAS
Given that it is possible, the following:
φ i + 1 ( N ) = HAS φ i + 1 ( N ) = HAS varphi_(i+1)(N)=A\varphi_{i+1}(\mathrm{~N})=\mathrm{A}φi+1( N)=HAS
is impossible, we will say that A can be an indicator of order i.
If equation (2) is possible for all i i iii, A can be an indicator of any order.
The formula:
(3)
φ ( 2 α + 1 ) = 2 α φ 2 α + 1 = 2 α varphi(2^(alpha+1))=2^(alpha)\varphi\left(2^{\alpha+1}\right)=2^{\alpha}φ(2α+1)=2α
Let's take:
φ 1 ( N ) = φ ( N ) , φ 2 ( N ) = φ ( φ 1 ( N ) ) , φ 3 ( N ) = φ ( φ 2 ( N ) ) , etc. φ 1 ( N ) = φ ( N ) , φ 2 ( N ) = φ φ 1 ( N ) , φ 3 ( N ) = φ φ 2 ( N ) ,  etc.  varphi_(1)(N)=varphi(N),quadvarphi_(2)(N)=varphi(varphi_(1)((N))),quadvarphi_(3)(N)=varphi(varphi_(2)((N))),quad" etc."\varphi_{1}(\mathrm{~N})=\varphi(\mathrm{N}), \quad \varphi_{2}(\mathrm{~N})=\varphi\left(\varphi_{1}(\mathrm{~N})\right), \quad \varphi_{3}(\mathrm{~N})=\varphi\left(\varphi_{2}(\mathrm{~N})\right), \quad \text { etc. }φ1( N)=φ(N),φ2( N)=φ(φ1( N)),φ3( N)=φ(φ2( N)), etc. 
φ i ( N ) φ i ( N ) varphi_(i)(N)\varphi_{i}(\mathrm{~N})φi( N)will be the iterated indicator or simply the indicator of order i of
N. 1 0 1 0 1^(0)1^{0}10And 2 0 2 0 2^(0)2^{0}20it follows that there exists a number n n nnndependent on N such that:
φ n ( N ) = 1 1 ) φ n ( N ) = 1 1 ) varphi_(n)(N)=1^(1))\varphi_{n}(\mathrm{~N})=1^{1)}φn( N)=11)
φ n ( N ) φ n ( N ) varphi_(n)(N)\varphi_{n}(\mathrm{~N})φn( N)is then the last indicator of N.
This shows us that powers of 2 can be indicators of any order. Similarly, the equality:
(4) φ ( 2 α 3 β − 1 − 1 ) = 2 α 3 β (4) φ 2 α 3 β − 1 − 1 = 2 α 3 β {:(4)varphi(2^(alpha)3^(beta-1-1))=2^(alpha)3^(beta):}\begin{equation*} \varphi\left(2^{\alpha} 3^{\beta-1-1}\right)=2^{\alpha} 3^{\beta} \tag{4} \end{equation*}(4)φ(2α3β−1−1)=2α3β
proves to us that numbers of the form 2 α 3 β ( a ≠ 0 ) 2 α 3 β ( a ≠ 0 ) 2^(alpha)3^(beta)(a!=0)2^{\alpha} 3^{\beta}(a \neq 0)2α3β(has≠0)can be indicators of any order.
We propose to demonstrate that:
The only numbers that can be indicators of any order are numbers of the form:
(5) 2 α , ou 2 α 3 β , avec a ≠ 0 (5) 2 α ,  ou  2 α 3 β ,  avec  a ≠ 0 {:(5)2^(alpha)","quad" ou "2^(alpha)3^(beta)","quad" avec "quad a!=0:}\begin{equation*} 2^{\alpha}, \quad \text { ou } 2^{\alpha} 3^{\beta}, \quad \text { avec } \quad a \neq 0 \tag{5} \end{equation*}(5)2α, Or 2α3β, with has≠0
We need only examine the even numbers that are not of the form (5). Before continuing, note that if:
φ ( N ) = 2 α A φ ( N ) = 2 α A varphi(N)=2^(alpha)A\varphi(\mathrm{N})=2^{\alpha} \mathrm{A}φ(N)=2αHAS
A being odd and not a power of 3 ( A = 1 = 3 0 A = 1 = 3 0 A=1=3^(0)\mathrm{A}=1=3^{0}HAS=1=30is a power of 3), the number N cannot be of any of the forms (5). We will always keep this remark in mind.
3. Lemma. Let p be a positive integer, in what follows:
p 1 = 2 p + 1 , p 2 = 2 p 1 + 1 , p 3 = 2 p 2 + 1 , … p 1 = 2 p + 1 , p 2 = 2 p 1 + 1 , p 3 = 2 p 2 + 1 , … p_(1)=2p+1,quadp_(2)=2p_(1)+1,quadp_(3)=2p_(2)+1,dotsp_{1}=2 p+1, \quad p_{2}=2 p_{1}+1, \quad p_{3}=2 p_{2}+1, \ldotsp1=2p+1,p2=2p1+1,p3=2p2+1,…
he y y yyyhas at least one non-prime number.
Indeed:
1 0 1 0 1^(0)1^{0}10. If :
p ≡ 0 ( mod 3 ) p ≡ 0 ( mod 3 ) p-=0quad(mod3)p \equiv 0 \quad(\bmod 3)p≡0(mod3)
we have:
p 1 ≡ 1 ( mod 3 ) p 2 ≡ 0 ( mod 3 ) p 1 ≡ 1      ( mod 3 ) p 2 ≡ 0      ( mod 3 ) {:[p_(1)-=1,(mod3)],[p_(2)-=0,(mod3)]:}\begin{array}{ll} p_{1} \equiv 1 & (\bmod 3) \\ p_{2} \equiv 0 & (\bmod 3) \end{array}p1≡1(mod3)p2≡0(mod3)
and it is evident that p 2 > 3 p 2 > 3 p_(2) > 3p_{2}>3p2>3.
2 ∘ ⋅ Si : 2 ∘ ⋅ Si : 2^(@)*Si:2^{\circ} \cdot \mathrm{Si}:2∘⋅If:
p ≡ 1 p ≡ 1 p-=1p \equiv 1p≡1
(mod. 3),
we have:
(mod.3) p 1 ≡ 0 (mod.3) p 1 ≡ 0 {:(mod.3)p_(1)-=0:}\begin{equation*} p_{1} \equiv 0 \tag{mod.3} \end{equation*}(mod.3)p1≡0
unless p p pppnot equal to 1. In this case p 1 = 3 p 1 = 3 p_(1)=3p_{1}=3p1=3And:
p 8 ≡ 0 ( mod 3 ) p 8 ≡ 0 ( mod 3 ) p_(8)-=0quad(mod3)p_{8} \equiv 0 \quad(\bmod 3)p8≡0(mod3)
  1. If :
(mod.3) p ≡ − 1 (mod.3) p ≡ − 1 {:(mod.3)p-=-1:}\begin{equation*} p \equiv-1 \tag{mod.3} \end{equation*}(mod.3)p≡−1
we can ask:
p = 3 k − 1 p = 3 k − 1 p=3k-1p=3 k-1p=3k−1
So what?
p 1 = 2 ⋅ 3 ⋅ k − 1 p 2 = 2 2 ⋅ 3 ⋅ k − 1 p 3 = 2 3 ⋅ 3 ⋅ k − 1 p 1 = 2 ⋅ 3 ⋅ k − 1 p 2 = 2 2 ⋅ 3 ⋅ k − 1 p 3 = 2 3 ⋅ 3 ⋅ k − 1 {:[p_(1)=2*3*k-1],[p_(2)=2^(2)*3*k-1],[p_(3)=2^(3)*3*k-1]:}\begin{aligned} & p_{1}=2 \cdot 3 \cdot k-1 \\ & p_{2}=2^{2} \cdot 3 \cdot k-1 \\ & p_{3}=2^{3} \cdot 3 \cdot k-1 \end{aligned}p1=2⋅3⋅k−1p2=22⋅3⋅k−1p3=23⋅3⋅k−1

Let's state:

3 k − 1 = 2 m ⋅ k 1 3 k − 1 = 2 m ⋅ k 1 3k-1=2^(m)*k_(1)3 k-1=2^{m} \cdot k_{1}3k−1=2m⋅k1
k 1 k 1 k_(1)k_{1}k1being odd, then of congruence:
2 φ ( k 1 ) ≡ 1 2 φ k 1 ≡ 1 2^(varphi(k_(1)))-=12^{\varphi\left(k_{1}\right)} \equiv 12φ(k1)≡1
we deduce:
(1) 2 φ ( k 1 ) 3 ⋅ h ≡ 2 m k 1 + 1 ≡ 1 (1) 2 φ k 1 3 ⋅ h ≡ 2 m k 1 + 1 ≡ 1 {:(1)2^(varphi(k_(1)))3*h-=2^(m)k_(1)+1-=1:}\begin{equation*} 2^{\varphi\left(k_{1}\right)} 3 \cdot h \equiv 2^{m} k_{1}+1 \equiv 1 \tag{1} \end{equation*}(1)2φ(k1)3⋅h≡2mk1+1≡1
SO p φ ( k 1 ) p φ k 1 p_(varphi(k_(1)))p_{\varphi\left(k_{1}\right)}pφ(k1)is divisible by k 1 k 1 k_(1)k_{1}k1which is certainly smaller than him. The reasoning is flawed if k 1 = 1 k 1 = 1 k_(1)=1k_{1}=1k1=1. In this case:
3 k − 1 = 2 m = 2 2 n + 1 3 k − 1 = 2 m = 2 2 n + 1 3k-1=2^(m)=2^(2n+1)3 k-1=2^{m}=2^{2 n+1}3k−1=2m=22n+1
and we have:
2 2 n ⋅ 3 k ≡ 1 ( mod 5 ) n impair 2 2 n + 5 ⋅ 3 k ≡ 1 ( mod 5 ) n pair 2 2 n ⋅ 3 k ≡ 1      ( mod 5 )      n  impair  2 2 n + 5 ⋅ 3 k ≡ 1      ( mod 5 )      n  pair  {:[2^(2n)*3k-=1,(mod5),n" impair "],[2^(2n+5)*3k-=1,(mod5),n" pair "]:}\begin{array}{lll} 2^{2 n} \cdot 3 k \equiv 1 & (\bmod 5) & n \text { impair } \\ 2^{2 n+5} \cdot 3 k \equiv 1 & (\bmod 5) & n \text { pair } \end{array}22n⋅3k≡1(mod5)n odd 22n+5⋅3k≡1(mod5)n peer 
The lemma is therefore proven.
4. We will now prove the property, first for the case of a number of the form:

2 A

(Odd) ≠ 3 p ≠ 3 p !=3^(p)\neq 3^{p}≠3p).
We divide the demonstration into three parts.
I. A number of the form:
2 p 2 p 2p2 p2p
p being prime > 3 > 3 > 3>3>3cannot be an indicator of just any order.
From the equation:
φ ( N ) = 2 p φ ( N ) = 2 p varphi(N)=2p\varphi(\mathrm{N})=2 pφ(N)=2p
We deduce:
N = 2 q m N = 2 q m N=2q^(m)\mathrm{N}=2 q^{m}N=2qm
( q q qqqfirst)
hence:
φ ( 2 q m ) = q m − 1 ( q − 1 ) = 2 p φ 2 q m = q m − 1 ( q − 1 ) = 2 p varphi(2q^(m))=q^(m-1)(q-1)=2p\varphi\left(2 q^{m}\right)=q^{m-1}(q-1)=2 pφ(2qm)=qm−1(q−1)=2p
SO:
1 0 . m = 2 , q = p = 3 1 0 . m = 2 , q = p = 3 1^(0).quad m=2,quad q=p=3quad1^{0} . \quad m=2, \quad q=p=3 \quad10.m=2,q=p=3impossible by hypothesis.
2 ∘ . m = 1 , q = 2 p + 1 = p 1 2 ∘ . m = 1 , q = 2 p + 1 = p 1 2^(@).quad m=1,quad q=2p+1=p_(1)2^{\circ} . \quad m=1, \quad q=2 p+1=p_{1}2∘.m=1,q=2p+1=p1.
If p 1 p 1 p_(1)p_{1}p1is not first, 2 p 2 p 2p2 p2pcannot be an indicator. Otherwise:
φ ( 2 p 1 ) = 2 p φ 2 p 1 = 2 p varphi(2p_(1))=2p\varphi\left(2 p_{1}\right)=2 pφ(2p1)=2p
If p 2 = 2 p 1 + 1 p 2 = 2 p 1 + 1 p_(2)=2p_(1)+1p_{2}=2 p_{1}+1p2=2p1+1is not first, 2 p 1 2 p 1 2p_(1)2 p_{1}2p1therefore cannot be an indicator 2 p 2 p 2p2 p2pIt cannot be a second-order indicator. Otherwise.
φ 2 ( 2 p 2 ) = φ 1 ( 2 p 1 ) = 2 p φ 2 2 p 2 = φ 1 2 p 1 = 2 p varphi_(2)(2p_(2))=varphi_(1)(2p_(1))=2p\varphi_{2}\left(2 p_{2}\right)=\varphi_{1}\left(2 p_{1}\right)=2 pφ2(2p2)=φ1(2p1)=2p
We finally see that in order for 2 p 2 p 2p2 p2peither an order indicator i i iiiand not of order i + 1 i + 1 i+1i+1i+1The numbers must be:
p 1 = 2 p + 1 , p 2 = 2 p 1 + 1 , … p i = 2 p i − 1 + 1 p 1 = 2 p + 1 , p 2 = 2 p 1 + 1 , … p i = 2 p i − 1 + 1 p_(1)=2p+1,quadp_(2)=2p_(1)+1,dotsquadp_(i)=2p_(i-1)+1p_{1}=2 p+1, \quad p_{2}=2 p_{1}+1, \ldots \quad p_{i}=2 p_{i-1}+1p1=2p+1,p2=2p1+1,…pi=2pi−1+1
are first and that 2 p i + 1 2 p i + 1 2p_(i)+12 p_{i}+12pi+1not be prime. The property in question then follows from the proven lemma.
II. A number of the form:
(a>1) 2 p ℓ (a>1) 2 p ℓ {:(a>1)2p^(ℓ):}\begin{equation*} 2 p^{\ell} \tag{a>1} \end{equation*}(a>1)2pℓ
p p pppbeing first > 3 > 3 > 3>3>3cannot be an indicator of any order.

The equation:

φ ( N ) = 2 p α φ ( N ) = 2 p α varphi(N)=2p^(alpha)\varphi(\mathrm{N})=2 p^{\alpha}φ(N)=2pα
gives again:
N = 2 q ′ ′ ′ N = 2 q ′ ′ ′ N=2q^(''')\mathrm{N}=2 q^{\prime \prime \prime}N=2q′′′
(q first)
and:
q m − 1 ( q − 1 ) = 2 p e q m − 1 ( q − 1 ) = 2 p e q^(m-1)(q-1)=2p^(e)q^{m-1}(q-1)=2 p^{e}qm−1(q−1)=2pe
SO.
1 0 m = a + 1 , q = p = 3 1 0 m = a + 1 , q = p = 3 1^(0)m=a+1,quad q=p=31^{0} m=a+1, \quad q=p=310m=has+1,q=p=3impossible by hypothesis.
2 0 m = 1 , q = 2 p ′ ′ + 1 = p 1 2 0 m = 1 , q = 2 p ′ ′ + 1 = p 1 2^(0)quad m=1,quad q=2p^('')+1=p_(1)2^{0} \quad m=1, \quad q=2 p^{\prime \prime}+1=p_{1}20m=1,q=2p′′+1=p1.
If p 1 p 1 p_(1)p_{1}p1is not first, 2 p α 2 p α 2p^(alpha)2 p^{\alpha}2pαcannot be an indicator. Otherwise:
f ( 2 p 1 ) = 2 p a f 2 p 1 = 2 p a f(2p_(1))=2p^(a)f\left(2 p_{1}\right)=2 p^{a}f(2p1)=2phas
and the 2 p 1 2 p 1 2p_(1)2 p_{1}2p1is of a form already studied. Property II then follows from property I.
III. A number of the form:
2 p 1 u 1 − p 2 u 2 ⋯ p v u v ( p 1 < p 2 < … < p v , v > 1 ) 2 p 1 u 1 − p 2 u 2 ⋯ p v u v p 1 < p 2 < … < p v , v > 1 2p_(1)^(u_(1))-p_(2)^(u_(2))cdotsp_(v)^(u_(v))quad(p_(1) < p_(2) < dots < p_(v),v > 1)2 p_{1}^{u_{1}}-p_{2}^{u_{2}} \cdots p_{v}^{u_{v}} \quad\left(p_{1}<p_{2}<\ldots<p_{v}, v>1\right)2p1u1−p2u2⋯pvuv(p1<p2<…<pv,v>1)
p 1 , p 2 , … p 0 p 1 , p 2 , … p 0 p_(1),p_(2),dotsp_(0)p_{1}, p_{2}, \ldots p_{0}p1,p2,…p0being prime, cannot be an indicator of any order.
We must always have:
N = 2 q m N = 2 q m N=2q^(m)\mathrm{N}=2 q^{m}N=2qm
(because if N contained several odd prime factors, φ ( N ) φ ( N ) varphi(N)\varphi(\mathrm{N})φ(N)would be divisible at least purely 2 2 2 2 2^(2)2^{2}22).
Therefore, we have:
q m − 1 ( q − 1 ) = 2 p 1 α 1 p 2 α 2 … p v α n q m − 1 ( q − 1 ) = 2 p 1 α 1 p 2 α 2 … p v α n q^(m-1)(q-1)=2p_(1)^(alpha_(1))p_(2)^(alpha_(2))dotsp_(v)^(alpha_(n))q^{m-1}(q-1)=2 p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{v}^{\alpha_{n}}qm−1(q−1)=2p1α1p2α2…pvαn
1 0 1 0 1^(0)1^{0}10It is obvious that one can never have q = p 1 , p 2 , … q = p 1 , p 2 , … q=p_(1),p_(2),dotsq=p_{1}, p_{2}, \ldotsq=p1,p2,…on p 0 − 1 p 0 − 1 p_(0-1)p_{0-1}p0−1.
2 0 2 0 2^(0)2^{0}20.
m = a v + 1 q = p v = 2 p 1 a 1 p 2 a 2 … p v − 1 a v − 1 + 1 m = a v + 1 q = p v = 2 p 1 a 1 p 2 a 2 … p v − 1 a v − 1 + 1 m=a_(v)+1quad q=p_(v)=2p_(1)^(a_(1))p_(2)^(a_(2))dotsp_(v-1)^(a_(v)-1)+1m=a_{v}+1 \quad q=p_{v}=2 p_{1}^{a_{1}} p_{2}^{a_{2}} \ldots p_{v-1}^{a_{v}-1}+1m=hasv+1q=pv=2p1has1p2has2…pv−1hasv−1+1
So either this equality does not exist, or we have:
φ ( 2 p v α v + 1 ) = 2 p 1 α 1 p 2 α 2 ⋯ p v α v φ 2 p v α v + 1 = 2 p 1 α 1 p 2 α 2 ⋯ p v α v varphi(2p_(v)^(alpha_(v))+1)=2p_(1)^(alpha_(1))p_(2)^(alpha_(2))cdotsp_(v)^(alpha_(v))\varphi\left(2 p_{v}^{\alpha_{v}}+1\right)=2 p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{v}^{\alpha_{v}}φ(2pvαv+1)=2p1α1p2α2⋯pvαv
and the 2 p v α v + 1 2 p v α v + 1 2p_(v)^(alpha_(v)+1)2 p_{v}^{\alpha_{v}+1}2pvαv+1is of a form already studied. Property III follows from property II.
3 0 3 0 3^(0)3^{0}30.
m = 1 , 4 = 2 p 1 α 1 p 2 α 2 … p v α v + 1 = p ′ m = 1 , 4 = 2 p 1 α 1 p 2 α 2 … p v α v + 1 = p ′ m=1,quad4=2p_(1)^(alpha_(1))p_(2)^(alpha_(2))dotsp_(v)^(alpha_(v))+1=p^(')m=1, \quad 4=2 p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{v}^{\alpha_{v}}+1=p^{\prime}m=1,4=2p1α1p2α2…pvαv+1=p′
If p ′ p ′ p^(')p^{\prime}p′If the product is not prime, equality is impossible. Otherwise:
φ ( 2 p ′ ) = 2 p 1 α 1 p 2 α 2 ⋯ p v α v φ 2 p ′ = 2 p 1 α 1 p 2 α 2 ⋯ p v α v varphi(2p^('))=2p_(1)^(alpha_(1))p_(2)^(alpha_(2))cdotsp_(v)^(alpha_(v))\varphi\left(2 p^{\prime}\right)=2 p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{v}^{\alpha_{v}}φ(2p′)=2p1α1p2α2⋯pvαv
and property III then follows from property 1.
We can therefore state the following proposition:

A number of the form:

Since A is an odd number not a power of 3, it cannot be an indicator of any order.
5. We will now show that the numbers:
2 a A ( A impair ≠ 3 3 ) 2 a A  A impair  ≠ 3 3 2^(a)Aquad(" A impair "!=3^(3))2^{a} \mathrm{~A} \quad\left(\text { A impair } \neq 3^{3}\right)2has HAS( An odd number ≠33)
enjoy the same property. The proposition has been demonstrated for a = 1 a = 1 a=1a=1has=1It is therefore sufficient to prove that it remains true for a = k a = k a=ka=khas=kassuming it to be true u = 1 , 2 , … k − 1 u = 1 , 2 , … k − 1 u=1,2,dots k-1u=1,2, \ldots k-1u=1,2,…k−1We break down the proof as before for the case a = 1 a = 1 a=1a=1has=1IV
. A number of the form:
2 k p 2 k p 2^(k)p2^{k} p2kp
(p prime > 3)
cannot be an indicator of any order.
The equation:
φ ( N ) = 2 k p φ ( N ) = 2 k p varphi(N)=2^(k)p\varphi(\mathrm{N})=2^{k} pφ(N)=2kp
gives us:
(6)
N = 2 i p j q 1 q 2 … q 1 N = 2 i p j q 1 q 2 … q 1 N=2^(i)p^(j)q_(1)q_(2)dotsq_(1)\mathrm{N}=2^{i} p^{j} q_{1} q_{2} \ldots q_{1}N=2ipjq1q2…q1
( q 1 , q 2 , … q r q 1 , q 2 , … q r q_(1),q_(2),dotsq_(r)q_{1}, q_{2}, \ldots q_{r}q1,q2,…qrfirst ≠ p ≠ p !=p\neq p≠p) and distinct from others,
hence:
q ( N ) = 2 i − 1 p j − 1 ( p − 1 ) ( q 1 − 1 ) ( q 2 − 1 ) … ( q r − 1 ) j ≠ 0 p ( N ) = 2 i − 1 ( q 1 − 1 ) ( q 2 − 1 ) … ( q r − 1 ) j = 0 q ( N ) = 2 i − 1 p j − 1 ( p − 1 ) q 1 − 1 q 2 − 1 … q r − 1      j ≠ 0 p ( N ) = 2 i − 1 q 1 − 1 q 2 − 1 … q r − 1      j = 0 {:[q(N)=2^(i-1)p^(j-1)(p-1)(q_(1)-1)(q_(2)-1)dots(q_(r)-1),j!=0],[p(N)=2^(i-1)(q_(1)-1)(q_(2)-1)dots(q_(r)-1),j=0]:}\begin{array}{ll} q(\mathrm{~N})=2^{i-1} p^{j-1}(p-1)\left(q_{1}-1\right)\left(q_{2}-1\right) \ldots\left(q_{r}-1\right) & j \neq 0 \\ p(\mathrm{~N})=2^{i-1}\left(q_{1}-1\right)\left(q_{2}-1\right) \ldots\left(q_{r}-1\right) & j=0 \end{array}q( N)=2i−1pj−1(p−1)(q1−1)(q2−1)…(qr−1)j≠0p( N)=2i−1(q1−1)(q2−1)…(qr−1)j=0
and it is necessary that:
(8)
p = 2 n + 1 q 1 = 2 n 1 p n 1 ′ + 1 q 2 = 2 n 2 p n 2 ′ + 1 ⋅ ⋅ ⋅ q r = 2 n ⋅ p n 1 ′ + 1 p = 2 n + 1 q 1 = 2 n 1 p n 1 ′ + 1 q 2 = 2 n 2 p n 2 ′ + 1 ⋅ ⋅ ⋅ q r = 2 n ⋅ p n 1 ′ + 1 {:[p=2^(n)+1],[q_(1)=2^(n_(1))p^(n_(1)^('))+1],[q_(2)=2^(n_(2))p^(n_(2)^('))+1],[***],[q_(r)=2^(n*)p^(n_(1)^('))+1]:}\begin{aligned} & p=2^{n}+1 \\ & q_{1}=2^{n_{1}} p^{n_{1}^{\prime}}+1 \\ & q_{2}=2^{n_{2}} p^{n_{2}^{\prime}}+1 \\ & \cdot \cdot \cdot \\ & q_{r}=2^{n \cdot} p^{n_{1}^{\prime}}+1 \end{aligned}p=2n+1q1=2n1pn1′+1q2=2n2pn2′+1⋅⋅⋅qr=2n⋅pn1′+1
1 ′ ′ 1 ′ ′ 1^('')1^{\prime \prime}1′′. If j ≠ 0 j ≠ 0 j!=0j \neq 0j≠0we have:
i − 1 + n + n 1 + n 2 + … + n r = k j − 1 + n 1 ′ + n 2 ′ + … + n r ′ = 1 i − 1 + n + n 1 + n 2 + … + n r = k j − 1 + n 1 ′ + n 2 ′ + … + n r ′ = 1 {:[i-1+n+n_(1)+n_(2)+dots+n_(r)=k],[j-1+n_(1)^(')+n_(2)^(')+dots+n_(r)^(')=1]:}\begin{aligned} & i-1+n+n_{1}+n_{2}+\ldots+n_{r}=k \\ & j-1+n_{1}^{\prime}+n_{2}^{\prime}+\ldots+n_{r}^{\prime}=1 \end{aligned}i−1+n+n1+n2+…+nr=kj−1+n1′+n2′+…+nr′=1
We can't have i > k i > k i > ki>ki>k. If i < k i < k i < ki<ki<kthe number N obtained falls into the case a < k a < k a < ka<khas<kfor which the property is true by hypothesis. If i < k i < k i < ki<ki<kwe must have:
n = 1 , n 1 + n 2 + … + n r = 0 ( n 1 = n 2 = … = n r = 0 ) n = 1 , n 1 + n 2 + … + n r = 0 n 1 = n 2 = … = n r = 0 n=1,quadn_(1)+n_(2)+dots+n_(r)=0quad(n_(1)=n_(2)=dots=n_(r)=0)n=1, \quad n_{1}+n_{2}+\ldots+n_{r}=0 \quad\left(n_{1}=n_{2}=\ldots=n_{r}=0\right)n=1,n1+n2+…+nr=0(n1=n2=…=nr=0)
that's to say N = 2 k p 2 N = 2 k p 2 N=2^(k)p^(2)N=2^{k} p^{2}N=2kp2And p = 3 p = 3 p=3p=3p=3which is excluded by hypothesis.
20. If j = 0 j = 0 j=0j=0j=0we have:
i − 1 + n 1 + n 2 + … + n r = k n 2 ′ + n 2 ′ + … + n r ′ = 1 i − 1 + n 1 + n 2 + … + n r = k n 2 ′ + n 2 ′ + … + n r ′ = 1 {:[i-1+n_(1)+n_(2)+dots+n_(r)=k],[n_(2)^(')+n_(2)^(')+dots+n_(r)^(')=1]:}\begin{gathered} i-1+n_{1}+n_{2}+\ldots+n_{r}=k \\ n_{2}^{\prime}+n_{2}^{\prime}+\ldots+n_{r}^{\prime}=1 \end{gathered}i−1+n1+n2+…+nr=kn2′+n2′+…+nr′=1
and the only case that needs to be studied is i = k i = k i=ki=ki=k, SO:
n 1 = 1 , n 2 ^ + … + n r = 0 , r = 1 n 1 ′ = 1 n 1 = 1 , n 2 ^ + … + n r = 0 , r = 1 n 1 ′ = 1 {:[n_(1)=1","quad hat(n_(2))+dots+n_(r)=0","quad r=1],[n_(1)^(')=1]:}\begin{gathered} n_{1}=1, \quad \hat{n_{2}}+\ldots+n_{r}=0, \quad r=1 \\ n_{1}^{\prime}=1 \end{gathered}n1=1,n2^+…+nr=0,r=1n1′=1
And :
φ ( 2 h ( 2 p + 1 ) ) = 2 h p φ 2 h ( 2 p + 1 ) = 2 h p varphi(2^(h)(2p+1))=2^(h)p\varphi\left(2^{h}(2 p+1)\right)=2^{h} pφ(2h(2p+1))=2hp
If p 1 = 2 p + 1 p 1 = 2 p + 1 p_(1)=2p+1p_{1}=2 p+1p1=2p+1If the product is not prime, equality is impossible. Otherwise 2 k ( 2 p + 1 ) = 2 k p 1 2 k ( 2 p + 1 ) = 2 k p 1 2^(k)(2p+1)=2^(k)p_(1)2^{k}(2 p+1)=2^{k} p_{1}2k(2p+1)=2kp1is of the same shape as 2 k p 2 k p 2^(k)p2^{k} p2kpIt is easy to see that everything boils down to showing that the following:
p 1 = 2 p − 1 , p 2 = 2 p 1 + 1 , p 3 = 2 p 2 + 1 , … p 1 = 2 p − 1 , p 2 = 2 p 1 + 1 , p 3 = 2 p 2 + 1 , … p_(1)=2p-1,quadp_(2)=2p_(1)+1,quadp_(3)=2p_(2)+1,dotsp_{1}=2 p-1, \quad p_{2}=2 p_{1}+1, \quad p_{3}=2 p_{2}+1, \ldotsp1=2p−1,p2=2p1+1,p3=2p2+1,…
contains at least one non-prime number, which follows from the proven lemma.
V. A number of the form:
2 h p β ( p premier > 3 , β > 3 ) 2 h p β ( p  premier  > 3 , β > 3 ) 2^(h)p^(beta)quad(p" premier " > 3,beta > 3)2^{h} p^{\beta} \quad(p \text { premier }>3, \beta>3)2hpβ(p first >3,β>3)
cannot be an indicator of just any order.
We always have the form (6) of N with j = 0 , 1 , 2 , … β + 1 j = 0 , 1 , 2 , … β + 1 j=0,1,2,dots beta+1j=0,1,2, \ldots \beta+1j=0,1,2,…β+1We will therefore write relations (7) and (8).
1 ∘ 1 ∘ 1^(@)1^{\circ}1∘. If j ≠ 0 j ≠ 0 j!=0j \neq 0j≠0we have:
i − 1 + n + n 1 + … + n r = k j − 1 + n 1 ′ + n 2 ′ + … + n r ′ = β . i − 1 + n + n 1 + … + n r = k j − 1 + n 1 ′ + n 2 ′ + … + n r ′ = β . {:[i-1+n+n_(1)+dots+n_(r)=k],[j-1+n_(1)^(')+n_(2)^(')+dots+n_(r)^(')=beta.]:}\begin{aligned} & i-1+n+n_{1}+\ldots+n_{r}=k \\ & j-1+n_{1}^{\prime}+n_{2}^{\prime}+\ldots+n_{r}^{\prime}=\beta . \end{aligned}i−1+n+n1+…+nr=kj−1+n1′+n2′+…+nr′=β.
We can still see that the case i > k i > k i > ki>ki>kis impossible and that i < k i < k i < ki<ki<kreduces to a supposedly proven case. It therefore remains i = k i = k i=ki=ki=kas the last possible case. But in this case n = 1 n = 1 n=1n=1n=1And p = 3 p = 3 p=3p=3p=3, which is by definition impossible.
2 0 2 0 2^(0)2^{0}20. If j = 0 j = 0 j=0j=0j=0The only case to consider is i = k i = k i=ki=ki=k, SO :
n 1 = 1 , n 2 + n 3 + … + n r = 0 , r = 1 n 1 ′ = β n 1 = 1 , n 2 + n 3 + … + n r = 0 , r = 1 n 1 ′ = β {:[n_(1)=1","quadn_(2)+n_(3)+dots+n_(r)=0","quad r=1],[n_(1)^(')=beta]:}\begin{gathered} n_{1}=1, \quad n_{2}+n_{3}+\ldots+n_{r}=0, \quad r=1 \\ n_{1}^{\prime}=\beta \end{gathered}n1=1,n2+n3+…+nr=0,r=1n1′=β
And:
N = 2 k ( 2 p α + 1 ) N = 2 k 2 p α + 1 N=2^(k)(2p^(alpha)+1)\mathrm{N}=2^{k}\left(2 p^{\alpha}+1\right)N=2k(2pα+1)
assuming of course that 2 p c + 1 2 p c + 1 2p^(c)+12 p^{c}+12pc+1is prime. But the number 2 h ( 2 p c + 1 ) 2 h 2 p c + 1 2^(h)(2p^(c)+1)2^{h}\left(2 p^{c}+1\right)2h(2pc+1)taken as an indicator has already been studied and property V results from property IV.
Finally, we must demonstrate that:
VI. A number of the form:
2 h p 1 α 1 p 2 α 2 ⋯ p v α v ( p 1 , p 2 , … p v premiers, v > 1 ) 2 h p 1 α 1 p 2 α 2 ⋯ p v α v p 1 , p 2 , … p v  premiers,  v > 1 2^(h)p_(1)^(alpha_(1))p_(2)^(alpha_(2))cdotsp_(v)^(alpha_(v))quad(p_(1),p_(2),dotsp_(v)" premiers, "v > 1)2^{h} p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{v}^{\alpha_{v}} \quad\left(p_{1}, p_{2}, \ldots p_{v} \text { premiers, } v>1\right)2hp1α1p2α2⋯pvαv(p1,p2,…pv first, v>1)
cannot be an indicator of any order.
We must have:
N = 2 i p 1 j 1 p 2 j 2 … p v j v q 1 q 1 … q , ( q 1 , q 2 , … q , premiers distincts et distincts de p 1 , p 2 , … p v j 2 = 0 , 1 , … a λ + 1 , λ = 1 , 2 , … v ) N = 2 i p 1 j 1 p 2 j 2 … p v j v q 1 q 1 … q , q 1 , q 2 , … q , premiers distincts   et distincts de  p 1 , p 2 , … p v j 2 = 0 , 1 , … a λ + 1 , λ = 1 , 2 , … v N=2^(i)p_(1)^(j_(1))p_(2)^(j_(2))dotsp_(v)^(j_(v))q_(1)q_(1)dotsq_(,)quad([q_(1)","q_(2)","dotsq_(,)"premiers distincts "],[" et distincts de "p_(1)","p_(2)","dotsp_(v)],[j_(2)=0","1","dotsa_(lambda)+1","lambda=1","2","dots v])\mathrm{N}=2^{i} p_{1}^{j_{1}} p_{2}^{j_{2}} \ldots p_{v}^{j_{v}} q_{1} q_{1} \ldots q_{,} \quad\left(\begin{array}{l} q_{1}, q_{2}, \ldots q_{,} \text {premiers distincts } \\ \text { et distincts de } p_{1}, p_{2}, \ldots p_{v} \\ j_{2}=0,1, \ldots a_{\lambda}+1, \lambda=1,2, \ldots v \end{array}\right)N=2ip1j1p2j2…pvjvq1q1…q,(q1,q2,…q,first distinct  and distinct from p1,p2,…pvj2=0,1,…hasλ+1,λ=1,2,…v)
and we obtain:
p ( N ) = 2 i − 1 p s j s − 1 p s + 1 j s + 1 − 1 … p v j v − 1 ( p s − 1 ) ( p s + 1 − 1 ) … ( p v − 1 ) ( q 1 − 1 ) ( q 2 − 1 ) … ( q v − 1 ) ( j 1 = j 2 = … = j s − 1 = 0 , j s = j s + 1 , … j ≠ 0 p ( N ) = 2 i − 1 ( q 1 − 1 ) ( q 2 − 1 ) … ( q 1 − 1 ) ( j 1 = j 2 = … = j v = 0 ) . p ( N ) = 2 i − 1 p s j s − 1 p s + 1 j s + 1 − 1 … p v j v − 1 p s − 1 p s + 1 − 1 … p v − 1 q 1 − 1 q 2 − 1 … q v − 1 j 1 = j 2 = … = j s − 1 = 0 , j s = j s + 1 , … j ≠ 0 p ( N ) = 2 i − 1 q 1 − 1 q 2 − 1 … q 1 − 1 j 1 = j 2 = … = j v = 0 . {:[p(N)=2^(i-1)p_(s)^(j_(s)-1)p_(s+1)^(j_(s+1)-1)dotsp_(v)^(j_(v)-1)(p_(s)-1)(p_(s+1)-1)dots(p_(v)-1)],[(q_(1)-1)(q_(2)-1)dots(q_(v)-1)],[(j_(1)=j_(2)=dots=j_(s-1)=0,j_(s)=j_(s+1),dots j!=0:}],[p(N)=2^(i-1)(q_(1)-1)(q_(2)-1)dots(q_(1)-1)quad(j_(1)=j_(2)=dots=j_(v)=0).]:}\begin{array}{r} p(\mathrm{~N})=2^{i-1} p_{s}^{j_{s}-1} p_{s+1}^{j_{s+1}-1} \ldots p_{v}^{j_{v}-1}\left(p_{s}-1\right)\left(p_{s+1}-1\right) \ldots\left(p_{v}-1\right) \\ \left(q_{1}-1\right)\left(q_{2}-1\right) \ldots\left(q_{v}-1\right) \\ \left(j_{1}=j_{2}=\ldots=j_{s-1}=0, j_{s}=j_{s+1}, \ldots j \neq 0\right. \\ p(\mathrm{~N})=2^{i-1}\left(q_{1}-1\right)\left(q_{2}-1\right) \ldots\left(q_{1}-1\right) \quad\left(j_{1}=j_{2}=\ldots=j_{v}=0\right) . \end{array}p( N)=2i−1psjs−1ps+1js+1−1…pvjv−1(ps−1)(ps+1−1)…(pv−1)(q1−1)(q2−1)…(qv−1)(j1=j2=…=js−1=0,js=js+1,…j≠0p( N)=2i−1(q1−1)(q2−1)…(q1−1)(j1=j2=…=jv=0).
To satisfy these equalities, we must assume that:
p μ = 2 m μ B μ + 1 q μ = 2 n μ B μ ′ + 1 p μ = 2 m μ B μ + 1 q μ = 2 n μ B μ ′ + 1 {:[p_(mu)=2^(m_(mu))B_(mu)+1],[q_(mu)=2^(n_(mu))B_(mu)^(')+1]:}\begin{aligned} & p_{\mu}=2^{m_{\mu}} \mathrm{B}_{\mu}+1 \\ & q_{\mu}=2^{n_{\mu}} \mathrm{B}_{\mu}^{\prime}+1 \end{aligned}pμ=2mμBμ+1qμ=2nμBμ′+1
for all p μ p μ p_(mu)p_{\mu}pμwhich intervene in N as factors, and for q μ , μ = 1 q μ , μ = 1 q_(mu),mu=1q_{\mu}, \mu=1qμ,μ=1, 2 , … r , B u , B u ′ 2 , … r , B u , B u ′ 2,dots r,B_(u),B_(u)^(')2, \ldots r, \mathrm{~B}_{u}, \mathrm{~B}_{u}^{\prime}2,…r, Bu, Bu′are numbers of the form:
p v 1 n 1 ′ p v 2 n 2 ′ … q v s n s ′ p v 1 n 1 ′ p v 2 n 2 ′ … q v s n s ′ p_(v_(1))^(n_(1)^('))p_(v_(2))^(n_(2)^('))dotsq_(v_(s))^(n_(s)^('))p_{v_{1}}^{n_{1}^{\prime}} p_{v_{2}}^{n_{2}^{\prime}} \ldots q_{v_{s}}^{n_{s}^{\prime}}pv1n1′pv2n2′…qvsns′
v 1 , v 2 , … v s v 1 , v 2 , … v s v_(1),v_(2),dotsv_(s)v_{1}, v_{2}, \ldots v_{s}v1,v2,…vsbeing s s sssnumbers of the sequence 1 , 2 , … ν 1 , 2 , … ν 1,2,dots nu1,2, \ldots \nu1,2,…ν.
1 ′ ′ 1 ′ ′ 1^('')1^{\prime \prime}1′′. If j 1 = j 2 = … = j s − 1 = 0 , j s ≠ 0 , j s + 1 ≠ 0 , … j v ≠ 0 j 1 = j 2 = … = j s − 1 = 0 , j s ≠ 0 , j s + 1 ≠ 0 , … j v ≠ 0 j_(1)=j_(2)=dots=j_(s-1)=0,j_(s)!=0,j_(s+1)!=0,dotsj_(v)!=0j_{1}=j_{2}=\ldots=j_{s-1}=0, j_{s} \neq 0, j_{s+1} \neq 0, \ldots j_{v} \neq 0j1=j2=…=js−1=0,js≠0,js+1≠0,…jv≠0, We have :
i − 1 + ∑ μ = s v m μ + ∑ μ = 1 y n μ = k i − 1 + ∑ μ = s v   m μ + ∑ μ = 1 y   n μ = k i-1+sum_(mu=s)^(v)m_(mu)+sum_(mu=1)^(y)n_(mu)=ki-1+\sum_{\mu=s}^{v} m_{\mu}+\sum_{\mu=1}^{y} n_{\mu}=ki−1+∑μ=svmμ+∑μ=1ynμ=k
and the only case that requires study is i = k i = k i=ki=ki=kSo we deduce:
s = v , m v = 1 , n 1 + n 2 + … + n r = 0 s = v , m v = 1 , n 1 + n 2 + … + n r = 0 s=v,quadm_(v)=1,quadn_(1)+n_(2)+dots+n_(r)=0s=v, \quad m_{v}=1, \quad n_{1}+n_{2}+\ldots+n_{r}=0s=v,mv=1,n1+n2+…+nr=0
And:
N = 2 k P v a v + 1 N = 2 k P v a v + 1 N=2^(k)P_(v)^(a_(v)+1)\mathrm{N}=2^{k} \mathrm{P}_{v}^{a_{v}+1}N=2kPvhasv+1
with:
p v = 2 p 1 α 1 p 2 α 3 ⋯ p v − 1 α v − 1 + 1 p v = 2 p 1 α 1 p 2 α 3 ⋯ p v − 1 α v − 1 + 1 p_(v)=2p_(1)^(alpha_(1))p_(2)^(alpha_(3))cdotsp_(v-1)^(alpha_(v-1))+1p_{v}=2 p_{1}^{\alpha_{1}} p_{2}^{\alpha_{3}} \cdots p_{v-1}^{\alpha_{v-1}}+1pv=2p1α1p2α3⋯pv−1αv−1+1
If such an equality is possible (by interchanging the order of the factors if necessary in 2 k p 1 α 1 p 2 α 2 … p v α v 2 k p 1 α 1 p 2 α 2 … p v α v 2^(k)p_(1)^(alpha_(1))p_(2)^(alpha_(2))dotsp_(v)^(alpha_(v))2^{k} p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{v}^{\alpha_{v}}2kp1α1p2α2…pvαv), the number N falls into a category already studied and the property v 1 v 1 v_(1)v_{1}v1results from property V:
2 0 2 0 2^(0)2^{0}20. If j 1 = j 2 = … = j v = 0 j 1 = j 2 = … = j v = 0 j_(1)=j_(2)=dots=j_(v)=0j_{1}=j_{2}=\ldots=j_{v}=0j1=j2=…=jv=0, We have :
i − 1 + ∑ μ = 1 x n μ = k i − 1 + ∑ μ = 1 x   n μ = k i-1+sum_(mu=1)^(x)n_(mu)=ki-1+\sum_{\mu=1}^{x} n_{\mu}=ki−1+∑μ=1xnμ=k
done if i = k , r = 1 i = k , r = 1 i=k,r=1i=k, r=1i=k,r=1And n 1 = 1 n 1 = 1 n_(1)=1n_{1}=1n1=1The number N is of the form:
2 k q 2 k q 2^(k)q2^{k} q2kq
with:
q = 2 p 1 α 1 p 2 α 2 … p v α v + 1 q = 2 p 1 α 1 p 2 α 2 … p v α v + 1 q=2p_(1)^(alpha_(1))p_(2)^(alpha_(2))dotsp_(v)^(alpha_(v))+1q=2 p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{v}^{\alpha_{v}}+1q=2p1α1p2α2…pvαv+1
and when this equality is possible we are brought back to property IV.
We can therefore state:
Theorem. A number of the form:
( k ≠ 0 ) 2 k A ( k ≠ 0 ) 2 k A {:((k!=0)")"2^(k)A:}\begin{equation*} 2^{k} \mathrm{~A} \tag{$k\neq0$} \end{equation*}(k≠0)2k HAS
Since A is odd and not a power of 3, it cannot be an indicator of any order.
It follows that for such a number one can determine a i i iiisuch that the equation:
φ i ( N ) = 2 k A φ i ( N ) = 2 k A varphi_(i)(N)=2^(k)A\varphi_{i}(\mathrm{~N})=2^{k} \mathrm{~A}φi( N)=2k HAS
either possible and the following:
φ i + 1 ( N ) = 2 k A φ i + 1 ( N ) = 2 k A varphi_(i+1)(N)=2^(k)A\varphi_{i+1}(\mathrm{~N})=2^{k} \mathrm{~A}φi+1( N)=2k HAS
impossible.
The proof given for the stated theorem can be used as a calculation method for the number i i iii.
    1. We studied the number n n nnnin a short work which will soon appear in the "Gazeta Matematică".

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