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I need to prove the following: $$\begin{aligned} &\text { Prove that } a^{N-1} \not\equiv 1(\bmod N) \text { if } \operatorname{gcd}(a, N)>1, \text { where } a, N \in \mathbb{Z} \text { and }\\ &N \geq 1 \end{aligned}$$

My attempt on a solution. set $d=gcd(a,N)>1$. Since $d$ divides a, we have: $$a\equiv 0 (\bmod d)$$ using this repeatedly we have: $$a^{N-1}\equiv 0 (\bmod d)$$

Since $d\mid N$ does that imply: $$\implies a^{N-1}\equiv 0 (\bmod N)$$?

I'm not quite convinced that this is true. If $d\mid a^{N-1}$ and $d \mid N$. Does that imply $N \mid a^{N-1}$? And how should I use the assumption $d>1$?.

I believe it would be obvious if it was the other way around. if $N \mid d$

sjm23
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  • The implication doesn't hold, but try to derive a contradiction: what happens if $a^{N-1}\equiv 1\pmod N$? How does it contradict $a^{N-1}\equiv 0\pmod d$? – player3236 Sep 27 '20 at 12:14
  • Because of $\gcd(a,N)>1$, there must be a prime $p$ dividing $a$ and $N$. $$a^{N-1}\equiv 1\mod N$$ implies $$a^{N-1}\equiv 1\mod p$$ contradicting $$a^{N-1}\equiv 0\mod p$$ – Peter Sep 27 '20 at 12:18
  • You can also apply this for $d:=\gcd(a,N)>1$ instead of $p$. – Peter Sep 27 '20 at 12:21
  • @player3236 What is wrong with the following proof, where I don't really use $d>1$ for anything.

    Let $d=gcd(a,N)>1$. assume for contradiction that $a^{N-1}\equiv 1(\bmod N)$. Since $d\mid N$ this implies $$a^{N-1}\equiv 1(\bmod d)$$ which is a contradiction. Since $d\mid a$ implies $a\equiv 0 (\bmod d)$ which again implies $$a^{N-1}\equiv 0 (\bmod d) \not\equiv 1 (\bmod d)$$

     – sjm23 Sep 27 '20 at 13:03
  • If $d=1$ you could say that $0\equiv1\pmod d$. – player3236 Sep 27 '20 at 13:33
  • Special case $,x = a^{N-2},, b=1,$ in $,ax\equiv b\pmod{!N},$ is solvable $\iff \gcd(a,n)\mid b,,$ by the dupe. – Bill Dubuque Sep 29 '20 at 08:38

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Suppose, for the sake of contradiction, that, in fact, there exists some integer $k$ such that $$a^{N-1}=kN+1$$ Since $d\mid a\implies d\mid a^{N-1}$ and $d\mid N\implies d\mid kN$ we infer that $d\mid 1$, but this is absurd if $d>1$.

Dr. Mathva
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