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 N N N is 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)=14 φ(N) = 14 φ ( 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) φ i ( N ) = HAS Given A. It is evident that if equation (2) is possible for
i = m
i = m i=m i=m i = m , all equations (2), where
i = 1 , 2 , … m − 1
i = 1 , 2 , … m − 1 i=1,2,dots m-1 i=1,2, \ldots m-1 i = 1 , 2 , … m − 1 will be possible and if the equation with
i = m
i = m i=m i=m i = m is impossible, all equations with
i = m + 1
i = m + 1 i=m+1 i=m+1 i = m + 1 ,
m + 2 , …
m + 2 , … m+2,dots m+2, \ldots m + 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} φ i ( 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} φ i + 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 i i i , 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 ( 2 α + 1 ) = 2 α 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. } φ 1 ( N ) = φ ( N ) , φ 2 ( N ) = φ ( φ 1 ( N ) ) , φ 3 ( N ) = φ ( φ 2 ( N ) ) , etc.
φ
i
(
N
)
φ
i
(
N
) varphi_(i)(N) \varphi_{i}(\mathrm{~N}) φ i ( N ) will be the iterated indicator or simply the indicator of order i of
N.
1
0
1
0
1^(0) 1^{0} 1 0 And
2
0
2
0
2^(0) 2^{0} 2 0 it follows that there exists a number
n
n n n n dependent 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)} φ n ( N ) = 1 1 )
φ
n
(
N
)
φ
n
(
N
) varphi_(n)(N) \varphi_{n}(\mathrm{~N}) φ n ( 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 ( 2 α 3 β − 1 − 1 ) = 2 α 3 β
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) 2 α 3 β ( 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) 2 α , Or 2 α 3 β , 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 ) = 2 α HAS 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 = 3 0 is 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,dots p_{1}=2 p+1, \quad p_{2}=2 p_{1}+1, \quad p_{3}=2 p_{2}+1, \ldots p 1 = 2 p + 1 , p 2 = 2 p 1 + 1 , p 3 = 2 p 2 + 1 , … he
y
y
y y y has at least one non-prime number.
Indeed:
1
0
1
0
1^(0) 1^{0} 1 0 . If :
p
≡
0
(
mod
3
)
p
≡
0
(
mod
3
)
p-=0quad(mod3) p \equiv 0 \quad(\bmod 3) p ≡ 0 ( mod 3 ) 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} p 1 ≡ 1 ( mod 3 ) p 2 ≡ 0 ( mod 3 ) and it is evident that
p
2
>
3
p
2
>
3
p_(2) > 3 p_{2}>3 p 2 > 3 .
2
∘
⋅
Si
:
2
∘
⋅
Si
:
2^(@)*Si: 2^{\circ} \cdot \mathrm{Si}: 2 ∘ ⋅ If :
p
≡
1
p
≡
1
p-=1 p \equiv 1 p ≡ 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) p 1 ≡ 0 unless
p
p
p p p not equal to 1. In this case
p
1
=
3
p
1
=
3
p_(1)=3 p_{1}=3 p 1 = 3 And:
p
8
≡
0
(
mod
3
)
p
8
≡
0
(
mod
3
)
p_(8)-=0quad(mod3) p_{8} \equiv 0 \quad(\bmod 3) p 8 ≡ 0 ( mod 3 )
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-1 p=3 k-1 p = 3 k − 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} p 1 = 2 ⋅ 3 ⋅ k − 1 p 2 = 2 2 ⋅ 3 ⋅ k − 1 p 3 = 2 3 ⋅ 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} 3 k − 1 = 2 m ⋅ k 1
k
1
k
1
k_(1) k_{1} k 1 being odd, then of congruence:
2
φ
(
k
1
)
≡
1
2
φ
k
1
≡
1
2^(varphi(k_(1)))-=1 2^{\varphi\left(k_{1}\right)} \equiv 1 2 φ ( k 1 ) ≡ 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) 2 φ ( k 1 ) 3 ⋅ h ≡ 2 m k 1 + 1 ≡ 1 SO
p
φ
(
k
1
)
p
φ
k
1
p_(varphi(k_(1))) p_{\varphi\left(k_{1}\right)} p φ ( k 1 ) is divisible by
k
1
k
1
k_(1) k_{1} k 1 which is certainly smaller than him. The reasoning is flawed if
k
1
=
1
k
1
=
1
k_(1)=1 k_{1}=1 k 1 = 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} 3 k − 1 = 2 m = 2 2 n + 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} 2 2 n ⋅ 3 k ≡ 1 ( mod 5 ) n odd 2 2 n + 5 ⋅ 3 k ≡ 1 ( mod 5 ) 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} ≠ 3 p ).
We divide the demonstration into three parts.
I. A number of the form: p being prime
>
3
>
3
> 3 >3 > 3 cannot 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 p φ ( N ) = 2 p We deduce:
N
=
2
q
m
N
=
2
q
m
N=2q^(m) \mathrm{N}=2 q^{m} N = 2 q m
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 p φ ( 2 q m ) = q m − 1 ( q − 1 ) = 2 p SO:
1
0
.
m
=
2
,
q
=
p
=
3
1
0
.
m
=
2
,
q
=
p
=
3
1^(0).quad m=2,quad q=p=3quad 1^{0} . \quad m=2, \quad q=p=3 \quad 1 0 . m = 2 , q = p = 3 impossible 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 = 2 p + 1 = p 1 .
If
p
1
p
1
p_(1) p_{1} p 1 is not first,
2
p
2
p
2p 2 p 2 p cannot 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 p φ ( 2 p 1 ) = 2 p If
p
2
=
2
p
1
+
1
p
2
=
2
p
1
+
1
p_(2)=2p_(1)+1 p_{2}=2 p_{1}+1 p 2 = 2 p 1 + 1 is not first,
2
p
1
2
p
1
2p_(1) 2 p_{1} 2 p 1 therefore cannot be an indicator
2
p
2
p
2p 2 p 2 p It 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 p φ 2 ( 2 p 2 ) = φ 1 ( 2 p 1 ) = 2 p We finally see that in order for
2
p
2
p
2p 2 p 2 p either an order indicator
i
i
i i i and not of order
i
+
1
i
+
1
i+1 i+1 i + 1 The 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)+1 p_{1}=2 p+1, \quad p_{2}=2 p_{1}+1, \ldots \quad 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 are first and that
2
p
i
+
1
2
p
i
+
1
2p_(i)+1 2 p_{i}+1 2 p i + 1 not 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) 2 p ℓ
p
p
p p p being first
>
3
>
3
> 3 >3 > 3 cannot 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 ) = 2 p α gives again:
N
=
2
q
′
′
′
N
=
2
q
′
′
′
N=2q^(''') \mathrm{N}=2 q^{\prime \prime \prime} N = 2 q ′ ′ ′ (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} q m − 1 ( q − 1 ) = 2 p e 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=3 1^{0} m=a+1, \quad q=p=3 1 0 m = has + 1 , q = p = 3 impossible 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} 2 0 m = 1 , q = 2 p ′ ′ + 1 = p 1 .
If
p
1
p
1
p_(1) p_{1} p 1 is not first,
2
p
α
2
p
α
2p^(alpha) 2 p^{\alpha} 2 p α cannot 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 ( 2 p 1 ) = 2 p has and the
2
p
1
2
p
1
2p_(1) 2 p_{1} 2 p 1 is 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) 2 p 1 u 1 − p 2 u 2 ⋯ p v u v ( p 1 < p 2 < … < p v , 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} p 1 , p 2 , … p 0 being 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 = 2 q m (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} 2 2 ). 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}} q m − 1 ( q − 1 ) = 2 p 1 α 1 p 2 α 2 … p v α n
1
0
1
0
1^(0) 1^{0} 1 0 It is obvious that one can never have
q
=
p
1
,
p
2
,
…
q
=
p
1
,
p
2
,
…
q=p_(1),p_(2),dots q=p_{1}, p_{2}, \ldots q = p 1 , p 2 , … on
p
0
−
1
p
0
−
1
p_(0-1) p_{0-1} p 0 − 1 .
2
0
2
0
2^(0) 2^{0} 2 0 .
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)+1 m=a_{v}+1 \quad q=p_{v}=2 p_{1}^{a_{1}} p_{2}^{a_{2}} \ldots p_{v-1}^{a_{v}-1}+1 m = has v + 1 q = p v = 2 p 1 has 1 p 2 has 2 … p v − 1 has v − 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 ( 2 p v α v + 1 ) = 2 p 1 α 1 p 2 α 2 ⋯ p v α v 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} 2 p v α v + 1 is of a form already studied. Property III follows from property II.
3
0
3
0
3^(0) 3^{0} 3 0 .
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 = 2 p 1 α 1 p 2 α 2 … p v α v + 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 ( 2 p ′ ) = 2 p 1 α 1 p 2 α 2 ⋯ p v α v and property III then follows from property 1.
We can therefore state the following proposition:
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) 2 has HAS ( An odd number ≠ 3 3 ) enjoy the same property. The proposition has been demonstrated for
a
=
1
a
=
1
a=1 a=1 has = 1 It is therefore sufficient to prove that it remains true for
a
=
k
a
=
k
a=k a=k has = k assuming it to be true
u
=
1
,
2
,
…
k
−
1
u
=
1
,
2
,
…
k
−
1
u=1,2,dots k-1 u=1,2, \ldots k-1 u = 1 , 2 , … k − 1 We break down the proof as before for the case
a
=
1
a
=
1
a=1 a=1 has = 1 IV
. A number of the form:
2
k
p
2
k
p
2^(k)p 2^{k} p 2 k p (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} p φ ( N ) = 2 k p 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 = 2 i p j q 1 q 2 … q 1 (
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} q 1 , q 2 , … q r first
≠
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 ) = 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 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 = 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
1
′
′
1
′
′
1^('') 1^{\prime \prime} 1 ′ ′ . If
j
≠
0
j
≠
0
j!=0 j \neq 0 j ≠ 0 we 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 + n 1 + n 2 + … + n r = k j − 1 + n 1 ′ + n 2 ′ + … + n r ′ = 1 We can't have
i
>
k
i
>
k
i > k i>k i > k . If
i
<
k
i
<
k
i < k i<k i < k the number N obtained falls into the case
a
<
k
a
<
k
a < k a<k has < k for which the property is true by hypothesis. If
i
<
k
i
<
k
i < k i<k i < k we 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 , n 1 + n 2 + … + n r = 0 ( n 1 = n 2 = … = n r = 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 = 2 k p 2 And
p
=
3
p
=
3
p=3 p=3 p = 3 which is excluded by hypothesis.
20. If
j
=
0
j
=
0
j=0 j=0 j = 0 we 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 + n 1 + n 2 + … + n r = k n 2 ′ + n 2 ′ + … + n r ′ = 1 and the only case that needs to be studied is
i
=
k
i
=
k
i=k i=k i = 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} n 1 = 1 , n 2 ^ + … + n r = 0 , r = 1 n 1 ′ = 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} p φ ( 2 h ( 2 p + 1 ) ) = 2 h p If
p
1
=
2
p
+
1
p
1
=
2
p
+
1
p_(1)=2p+1 p_{1}=2 p+1 p 1 = 2 p + 1 If 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} 2 k ( 2 p + 1 ) = 2 k p 1 is of the same shape as
2
k
p
2
k
p
2^(k)p 2^{k} p 2 k p It 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,dots p_{1}=2 p-1, \quad p_{2}=2 p_{1}+1, \quad p_{3}=2 p_{2}+1, \ldots p 1 = 2 p − 1 , p 2 = 2 p 1 + 1 , p 3 = 2 p 2 + 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) 2 h p β ( 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+1 j=0,1,2, \ldots \beta+1 j = 0 , 1 , 2 , … β + 1 We will therefore write relations (7) and (8).
1
∘
1
∘
1^(@) 1^{\circ} 1 ∘ . If
j
≠
0
j
≠
0
j!=0 j \neq 0 j ≠ 0 we 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 + n 1 + … + n r = k j − 1 + n 1 ′ + n 2 ′ + … + n r ′ = β . We can still see that the case
i
>
k
i
>
k
i > k i>k i > k is impossible and that
i
<
k
i
<
k
i < k i<k i < k reduces to a supposedly proven case. It therefore remains
i
=
k
i
=
k
i=k i=k i = k as the last possible case. But in this case
n
=
1
n
=
1
n=1 n=1 n = 1 And
p
=
3
p
=
3
p=3 p=3 p = 3 , which is by definition impossible.
2
0
2
0
2^(0) 2^{0} 2 0 . If
j
=
0
j
=
0
j=0 j=0 j = 0 The only case to consider is
i
=
k
i
=
k
i=k i=k i = 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} n 1 = 1 , n 2 + n 3 + … + n r = 0 , r = 1 n 1 ′ = 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 = 2 k ( 2 p α + 1 ) assuming of course that
2
p
c
+
1
2
p
c
+
1
2p^(c)+1 2 p^{c}+1 2 p c + 1 is 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) 2 h ( 2 p c + 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) 2 h p 1 α 1 p 2 α 2 ⋯ p v α v ( p 1 , p 2 , … p v 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 = 2 i p 1 j 1 p 2 j 2 … p v j v q 1 q 1 … q , ( q 1 , q 2 , … q , first distinct and distinct from p 1 , p 2 , … p v j 2 = 0 , 1 , … has λ + 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 ) = 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 ) . 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} p μ = 2 m μ B μ + 1 q μ = 2 n μ B μ ′ + 1 for all
p
μ
p
μ
p_(mu) p_{\mu} p μ which intervene in N as factors, and for
q
μ
,
μ
=
1
q
μ
,
μ
=
1
q_(mu),mu=1 q_{\mu}, \mu=1 q μ , 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 , B u , B u ′ 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}} p v 1 n 1 ′ p v 2 n 2 ′ … q v s n s ′
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} v 1 , v 2 , … v s being
s
s
s s s numbers of the sequence
1
,
2
,
…
ν
1
,
2
,
…
ν
1,2,dots nu 1,2, \ldots \nu 1 , 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)!=0 j_{1}=j_{2}=\ldots=j_{s-1}=0, j_{s} \neq 0, j_{s+1} \neq 0, \ldots j_{v} \neq 0 j 1 = j 2 = … = j s − 1 = 0 , j s ≠ 0 , j s + 1 ≠ 0 , … j v ≠ 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)=k i-1+\sum_{\mu=s}^{v} m_{\mu}+\sum_{\mu=1}^{y} n_{\mu}=k i − 1 + ∑ μ = s v m μ + ∑ μ = 1 y n μ = k and the only case that requires study is
i
=
k
i
=
k
i=k i=k i = k So 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)=0 s=v, \quad m_{v}=1, \quad n_{1}+n_{2}+\ldots+n_{r}=0 s = v , m v = 1 , n 1 + n 2 + … + n r = 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 = 2 k P v has v + 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))+1 p_{v}=2 p_{1}^{\alpha_{1}} p_{2}^{\alpha_{3}} \cdots p_{v-1}^{\alpha_{v-1}}+1 p v = 2 p 1 α 1 p 2 α 3 ⋯ p v − 1 α v − 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}} 2 k p 1 α 1 p 2 α 2 … p v α v ), the number N falls into a category already studied and the property
v
1
v
1
v_(1) v_{1} v 1 results from property V:
2
0
2
0
2^(0) 2^{0} 2 0 . If
j
1
=
j
2
=
…
=
j
v
=
0
j
1
=
j
2
=
…
=
j
v
=
0
j_(1)=j_(2)=dots=j_(v)=0 j_{1}=j_{2}=\ldots=j_{v}=0 j 1 = j 2 = … = j v = 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)=k i-1+\sum_{\mu=1}^{x} n_{\mu}=k i − 1 + ∑ μ = 1 x n μ = k done if
i
=
k
,
r
=
1
i
=
k
,
r
=
1
i=k,r=1 i=k, r=1 i = k , r = 1 And
n
1
=
1
n
1
=
1
n_(1)=1 n_{1}=1 n 1 = 1 The number N is of the form:
2
k
q
2
k
q
2^(k)q 2^{k} q 2 k q 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))+1 q=2 p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{v}^{\alpha_{v}}+1 q = 2 p 1 α 1 p 2 α 2 … p v α v + 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 ) 2 k 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
i i i such 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} φ i ( N ) = 2 k 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} φ i + 1 ( N ) = 2 k HAS impossible.
The proof given for the stated theorem can be used as a calculation method for the number
i
i
i i i .
