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

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.

Use the formula $$x^p - y^p = (x - y)(x^{p - 1} + \dotsc + y^{p - 1})$$ and prove by induction.
• I am sorry but I do not get it yet. I tried to prove $x \equiv y \bmod p \implies x^{p^{k-1}} \equiv y^{p^{k-1}} \bmod {p^k}$ by induction as you suggested. For $k=1$ this is clear. But I do not see how the binomial theorem mentioned by you should help in the induction step. Could you please expand your hint a bit? – 3nondatur Nov 28 '19 at 10:33
• Prove by induction on $k$, and note that, if $x \equiv y \mod p$, then $x^{p-1}+\dotsc+y^{p-1}\equiv0\mod p$. These are all the ingredients you need. You just have to work a bit more on it. – WhatsUp Nov 28 '19 at 14:52