<s>[INST] Need to prove divisibility of $7^p-5^p-2$ by $3$ by using Fermat's little Theorem Let $p > 2$ be a prime number. Prove that $$7^p-5^p-2$$  can be divided by $6p$.
I have already proved that it can be divided by $2$ and $p$, but how can I prove that it can be divided by $3$ ?
 [/INST] Fermat's little theorem asserts that if $p$ is prime, then for every integer $n$, we have :
$$n^p-n\equiv 0\quad (mod\,p)$$
Since $5$ and $7$ are primes :
$$7^p-5^p-2=(7^p-7)-(5^p-5)\equiv0\quad(mod\,p)$$
But $p$ is also odd, so that the relation : $7^p-5^p-2\equiv1-(-1)^p-2\quad (mod\,6)$ takes the form :
$$7^p-5^p-2\equiv0\quad(mod\,6)$$
At this point, we have proven that if $p$ is an odd prime, then $p\mid(7^p-5^p-2)$ and $6\mid(7^p-5^p-2)$
If $p>3$, then $\gcd(p,6)=1$ and we conclude that $6p\mid(7^p-5^p-2)$.
And if $p=3$, then a direct calculation shows that $$7^3-5^3-2\equiv(7\times(-5))-5\times(7)-2=72\equiv0\quad(mod\,18)$$
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