1. ## Problem with Congruences

Let p is prime number.
Prove $a^p \equiv b^p\ (\mod p) \Rightarrow a^p \equiv b^p\ (\mod p^2)$. How to do it?

It is clear that: $a^p-b^p \equiv 0\ (\mod p)$

$a^p-b^p=(a-b)(a^{p-1}+a^{p-2}b+\ldots + b^{p-1})$, $a^{p-1}\equiv 1 (\mod p)$, $b^{p-1}\equiv 1 (\mod p)$

Is it $a^{p-i}b^{i-1}\equiv 1 (\mod p)$ and why?

Thank you.

2. ## Re: Problem with Congruences

Since $a^p\equiv a\,(mod\, p)$, $a=b+qp$ for some q. Thus
$$a^p-b^p=\sum_{k=1}^p{{p}\choose{k}}b^{p-k}(qp)^k=pb^{p-1}qp+\sum_{k=2}^p{{p}\choose{k}}b^{p-k}(qp)^k$$
We're done.
Worth remembering: $p$ divides each binomial coefficient ${{p}\choose{k}}\text{ for }1\leq k<p$.