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I came across several different proofs of Fermat's little theorem and in all of them what I ask to prove in this question is either used as an axiom(just a statement without any proofs) or just proven not very correctly(just an example of how this works with different numbers).

The main proof that I rely on is from an 8th grade textbook(which I currently learn algebra with), so I'm not sure, maybe it's not proven deliberately to free the proof of the Fermat's theorem from any redundant complications for 8th graders, but I can't understand the whole proof without understanding how to prove that when $ac ≡ bc (mod \space m)$, and $GCD(c, m) = 1$, then $a ≡ b (mod \space m)$.

I tried to prove it myself, but not sure if it works:

  1. $ac ≡ bc (mod \space m) \implies c(a - b) ≡ 0(mod \space m)$
  2. Since $GCD(c, m) = 1$ $\implies$ $c$ doesn't help to get a number that would divide $m$ $\implies$ $c(a - b) ≡ 0(mod \space m) \implies a ≡ b(mod \space m)$. And it's never true if $c$ and $m$ are not coprime.

Is this proof sufficient? If not, how to do it correctly?

curioushuman
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  • When (1) $c \not \equiv 0 \mod m$ & (2) $(a-b) \not \equiv 0 \mod m$ , then (3) $c(a-b) \not \equiv 0 \mod m$. When we have (4) $c(a-b) \equiv 0 \mod m$ with (1) , then we get the wanted Conclusion. – Prem May 29 '23 at 11:11

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