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Fermat's Little Theorem states that

If $p$ is a prime and $a \in Z$ with $gcd(a,p)=1$ then $a^{p-1} \equiv 1 \mod p$

from this I take to the converse to be the statement that

if $a^{p-1} \equiv 1 \mod p$ then $p$ is prime and $gcd(a,p)=1$

Taking the result stated here as a given, I make the claim that taking $p=(6t+1)(12t+1)(18t+1)$ for $t \in N$ disproves this because we have that $a^{p-1} \equiv 1 \mod p$ and yet $p$ is clearly not prime.

Is this a sufficient proof? I was informed that this answer was not sufficient in an assignment and I can't understand why?

cb7
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  • If you have proof that the particular $p$ you chose results in the stated congruence then yes, it is sufficient. Separately, a single counterexample is also considered sufficient. The giver of the assignment would have to clarify why your answer is insufficient. – abiessu Nov 27 '20 at 15:08
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    This answer is not appropriate, because an easy counterexample is expected, which you can justify yourself without copying Carmichael numbers (exactly what "lulu" says below in a better wording than mine :)) – Dietrich Burde Nov 27 '20 at 15:09
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    I would guess that the problem was that you quoted an unfamiliar result. And you can easily dodge that problem by giving an explicit example. – lulu Nov 27 '20 at 15:10
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    I think, you do not even need a Carmichael number. $2^{340}\equiv 1\mod 341$ would already qualify as a counterexample, unless you want to show that even if the congruence holds for all $a$ coprime to $p$ , $p$ need not be prime. Anyway, this is not a "not sufficient proof", whether it has a suitable level, is another story. – Peter Nov 27 '20 at 16:19
  • @lulu the previous question asked to prove that result so it's safe to assume we were supposed to use it somehow :) – cb7 Nov 27 '20 at 16:54
  • Note that there are ways to formulate little Fermat so that its converse is true, e.g. see here – Bill Dubuque Nov 27 '20 at 21:51

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