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I am reading a paper and they write that Fermat derived his little theorem from this observation: If $p$ is a prime, and $a$ is any number not divisible by $p$, then the order of a modulo $p$ divides $p−1$. In terms of divisiblity and this is the theorem that Fermat found the theorem looks like this:

If $p$ is prime and $a$ is coprime to p, then $p$ divides $a^{p−1}−1\implies p \mid a^{p-1}-1$

The multiplicative order in terms of divisibility looks like this: the order of $a$ mod $p$ is the smallest $e > 0$ so that $p$ divides $a^e−1\implies p \mid a^{e}-1$

I am trying to obtain Fermat's theorem by having a look on his observation: So I want to show that: "If $p$ is a prime, and $a$ is any number not divisible by $p$, then the order of a modulo $p$ divides $p−1$" leads to "If $p$ is prime and $a$ is coprime to p, then $p$ divides $a^{p−1}−1$".

My approach is, taking the definition of order: p | $a^e-1$ and using the observation $=>$p | $a^{e|p-1}-1$. This already looks pretty similar to $ p $ | $a^{p-1}-1$, but not quite and I am not sure if I am on the right track.

Thanks for your help

Bernard
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Banause
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  • Well note that any number coprime to $p$ that any power of it is coprime to $p$ and therefore by pigeonhole principle it will cycle at or before $p-1$ in exponent, as that's the number of coprime residues. – Roddy MacPhee Jun 19 '21 at 20:46
  • In the last paragraph I cannot make sense of $=>p|a^{e|p-1}-1$. BTW the LaTex for $\implies$ is \implies, and the LaTex for $\iff$ is \iff. – DanielWainfleet Jun 19 '21 at 21:32
  • You are right, maybe that does not make sense at all. – Banause Jun 19 '21 at 21:38
  • On second thought I'm not sure that's what you need. Are you trying to reproduce Fermat's original proof, or are you looking for any (modern) proof ? Do you know any group theory, esp. Lagrange's theorem – Bill Dubuque Jun 20 '21 at 00:30
  • In the paper they wrote that fermat observed the following: If p is a prime, and a is any number not divisible by p, then the order of a modulo p divides p−1. And I want to try with these given information to derive his theorem in terms of divisibility: $p | a^{p-1}-1$. I dont know if this is possible, I am just interested in how these definitions are related. My group theory knowledge is not that great. – Banause Jun 20 '21 at 07:27
  • It seems like you're having trouble with the definitions. The order, $e$ of an element, $a$ modulo $p$ is the smallest number such that $a^e \equiv 1 \pmod p$. It follows from the definition that for any multiple of $e$, $a^{ke} \equiv (a^e)^k \equiv 1 \pmod p$. Since you already know that $p-1$ is a multiple of $e$, we immediately get $a^{p-1} \equiv 1 \pmod p$. – arbashn Jun 20 '21 at 10:09
  • You are using the definitions in terms of congruence. But congruence was not a thing during Fermat's lifetime. That's why Fermat expressed his theorem in terms of divisibility. That's why I want to show the relation between the observation and the theorem Fermat found in terms of divisibility. Your idea still helps. I will try to use it, thanks. – Banause Jun 20 '21 at 10:38
  • Thanks arbashn your argumentation can be used the same way in terms of divisibility. So I understand the relation now. – Banause Jun 20 '21 at 12:46

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