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  1. Prove that if $\gcd(a,m)>1$, then $a^k\not\equiv1$ (mod $m$) for all $k\ge1$.

I know that the best way to do this is to prove it by contradiction. So, we say that $a^k \equiv 1$ (mod $m$). Also, we know that $a$ and $m$ are not relatively prime. I also feel like this proof would need Fermat’s little Theorem.

  1. Suppose $\gcd(a,m)=1$. Let $j$ be the order of $a$ modulo $m$. Prove that if $a^k\equiv1$ (mod $m$), then $j∣k$.
user729424
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ILUVU
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1 Answers1

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It's not really contradiction that you have to use, but rather a proof by contrapositive: show that if $a^k\equiv 1\pmod m$, then $\gcd(a,m)=1$.

For the second point, it has nothing to do with Fermat: it is a general result in group theory that if an element $g$ satisfies $g^k=e$ (the unit element of the groupˆ), then $k$ is a multiple of the order of $g$. It simply results from the group homomorphism \begin{align} (\mathbf Z,+)&\longrightarrow (G,\cdot), \\ k&\longmapsto g^k, \end{align} and $\mathbf Z$ being a P.I.D. (the order of $g$ is the positive generator of the kernel).

Bernard
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  • I understand your point, but for number 1, how would you do that. I know that from the definition of modulo, m divides a^k-1. What would I do from there? – ILUVU Apr 06 '20 at 23:27
  • Write $a^k=bm+1$ and conclude that any common divisor of $a$ and $m$ divides $1$. – Bernard Apr 06 '20 at 23:37