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Fom wiki Fermat's little theorem, the theorem is as follows: If there exists an integer $a$ such that $ a^{p-1}\equiv 1\pmod p $ and for all primes $q$ dividing $p − 1$ one has ${\displaystyle a^{(p-1)/q}\not \equiv 1{\pmod {p}},}$ then $p$ is prime.

How to prove this theorem?

miket
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  • What have you tried? – Thomas Andrews Feb 17 '23 at 06:39
  • @Thomas Andrews, I don't know were to begin, I tried search the net. – miket Feb 17 '23 at 06:44
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    Do you know about Euler's theorem? It says that $a^{\phi(n)} \equiv 1 ; (\textrm{mod} ; n)$ for all $a \in \mathbf{Z}$ where $\phi$ is Euler's totient function. Using this and the fact that $\phi(n) = n - 1$ if and only if $n$ is prime, you can prove your theorem! – Joseph Harrison Feb 17 '23 at 07:44
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    @JosephHarrison Yes I know Euler's theorem. – miket Feb 17 '23 at 08:37
  • @miket have you made any breakthroughs yet? – Joseph Harrison Feb 17 '23 at 08:53
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    @JosephHarrison Nothing made. – miket Feb 17 '23 at 12:42
  • @miket have you looked at the answer https://math.stackexchange.com/questions/26145/how-to-prove-lucass-converse-of-fermats-little-theorem-without-using-primitive ? – Joseph Harrison Feb 18 '23 at 16:18
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    @JosephHarrison Yes I looked at that answer, but it seems that answer doen't helps. – miket Feb 19 '23 at 02:57
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    @miket If $a^{p - 1} \equiv 1 \mod{p}$, then $\phi(p)$ divides $p - 1$. Writing $p - 1 = k\phi(p)$ for some integer $k$, if $k > 1$ then it is divisible by a prime $q$. Then $(p - 1)/q = k\phi(p)/q$, the latter of which is a multiple of $\phi(p)$ and so $a^{(p - 1)/q} \equiv 1 \mod{p}$, against assumption. Therefore we must have $k = 1$ so that $\phi(p) = p - 1$ and $p$ is prime. I hope this helps! Feel free to ask for clarifications. – Joseph Harrison Feb 19 '23 at 19:38
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    @JosephHarrison Thanks, I see. – miket Feb 21 '23 at 01:47
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    @miket awesome :) – Joseph Harrison Feb 21 '23 at 08:17

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