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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.

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

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  • $\begingroup$ 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? $\endgroup$ – 3nondatur Nov 28 '19 at 10:33
  • $\begingroup$ I made an edit to clarify my problem. $\endgroup$ – 3nondatur Nov 28 '19 at 10:49
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    $\begingroup$ 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. $\endgroup$ – WhatsUp Nov 28 '19 at 14:52

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