Let $p$ be prime. Prove if $p|(a^p - b^p)$, then $p^2 | (a^p - b^p)$

Question

Let $p$ be prime. Prove if $p|(a^p - b^p)$, then $p^2 | (a^p - b^p)$

don't know where to begin, any help appreciated

Answer 1

Hint: $a^p\cong b^p\pmod p\implies a\cong b\pmod p$ by Fermat's little theorem.

So you can write $a=b+k\cdot p$.

So, look at $(b+k\cdot p)^p-b^p$ and use the binomial theorem.

(Remember, $p\choose n$ is divisible by $p$ for $0\lt n\lt p$.)

Answer 2

Let p|(a^p - b^p)
Therefore p can be written as p=k*(a^p - b^p)
so p^2 becomes k^2*(a^p - b^p)^2
or, p^2=p*k*(a^p - b^p)
hence, p^ 2 |( a^ p − b^ p )