$x \equiv y \bmod p$ implies $x^{p^{k-1}} \equiv y^{p^{k-1}} \bmod {p^k}$?

Question

I am asking myself the following question:

Let $p$ be a prime. Is it true that $x \equiv y \bmod p$ implies $x^{p^{k-1}} \equiv y^{p^{k-1}} \bmod {p^k}$?

I can not see how to prove this, but I do not find a counter example either. Could you help me?

EDIT: We try to use induction on this one. Assume that $x \equiv y \bmod p$. We show that $x^{p^{k-1}} \equiv y^{p^{k-1}} \bmod {p^k}$ for all integers $k\ge 1$. The induction basis for $k=1$ is clear. So assume for all integers smaller than $k$ holds $x^{p^{k-1}} \equiv y^{p^{k-1}} \bmod {p^k}$. Now we do the induction step for $k+1$. So we need to show $x^{p^{k}} \equiv y^{p^{k}} \bmod {p^{k+1}}$. W can rewrite:

$$x^{p^{k}} - y^{p^{k}} = (x-y)(x^{p^{k-1}} + x^{p^{k-2}}y + \ldots + xy^{p^{k}-2} + y^{p^k-1})$$

By assmption we know that $x-y$ is a multiple of $p$ so we are left to show that

$$(x^{p^{k-1}} + x^{p^{k-2}}y + \ldots + xy^{p^{k}-2} + y^{p^k-1})$$

is a multiple of $p^{k}$. But I do not see why this sould be true.

Answer 1

Use the formula $x^p - y^p = (x - y)(x^{p - 1} + \dotsc + y^{p - 1})$ and prove by induction.