[SOLVED] Prove that a^p is congruent to b^p (mod p^2) if the same is true mod p

Quote:

For any prime $\displaystyle p$, if $\displaystyle a^p\equiv b^p\;(\text{mod } p)$, prove that $\displaystyle a^p\equiv b^p\;(\text{mod } p^2)$.

I have found this quite challenging!

I thought maybe we could divide it into two cases, where $\displaystyle p\big|a,b$ and $\displaystyle p\not\big|a,b$.

If we consider Euler's function $\displaystyle \phi(n)$, then we have, for $\displaystyle p\not\big|a,b$,

$\displaystyle b^{\phi(p^2)}\equiv a^{\phi(p^2)}\equiv 1\;(\text{mod }p^2)$.

Since $\displaystyle \phi(p^2)=p^2-p$, then

$\displaystyle b^{p^2-p}\equiv a^{p^2-p}\equiv 1\;(\text{mod }p^2)$.

So we could say that

$\displaystyle a^{p^2}\equiv a^p\;(\text{mod }p^2)$ and $\displaystyle b^{p^2}\equiv b^p\;(\text{mod }p^2)$.

And we can say a bunch of other things, too, none of which seem to get me closer to a solution.

Any help would be much appreciated!