Suppose $p \nmid b$ and $p \nmid a$. Let $a^p \equiv b^p (\text{mod}(p))$. Prove that $a^p \equiv b^p (\text{mod}(p^2))$.
Below is my proof. I'm interested in two things.
1) Verifing this proof
2) Finding a simpler proof (preferably one that doesn't involve the Binomial Theorem).
Suppose $a^p \equiv b^p (\text{mod}(p))$.
Since $a^{p - 1} \equiv b^{p - 1} \equiv 1 (\text{mod}(p))$ By Fermat's Little Theorem, we have $a^p \equiv a \equiv b^p \equiv b (\text{mod}(p))$.
Since $a \equiv b (\text{mod}(p))$ we can write $ a = pk + n$ and $b = pm + n$ where $k$ and $m$ are non-negative integers and $ 1 \leq n \leq p - 1$.
We then have $a^p = (pk + n)$ and $b^p = (pm + n)^p$.
Using the Binomial Theorem we can expand these. Looking at these expanded binomials modulo $p^2$ we see that any terms with powers of $p$ larger than 2 are 0 modulo $p^2$. Thus we are only concerned with the last two terms in the expansion. The expansions are displayed below:
$a^p = (pk + n)^p \equiv \frac{p!}{(p-1)!}n^{p-1}(pk) + n^p \equiv n^p\mod{p^2}$
and
$b^p = (pm + n)^p \equiv \frac{p!}{(p-1)!}n^{p-1}(pm) + n^p \equiv n^p\mod{p^2}$
Thus $a^p \equiv b^p \mod{p^2}$
I would really appreciate any critique on the proof including the formatting. Thanks.
