Congruence Modulo Powers of Primes

Question

I need help with the following:

Let $p$ be an odd prime and $x, y \in \mathbb{Z}$. Then $x \equiv y$ mod $p^k \implies x^p \equiv y^p$ mod $p^{k+1}$.

I know that I can write $x^p - y^p$ as $(x - y)F(x,y)$ for some ugly polynomial $F$, but I'm not sure whether that's too much help.

Ta

Answer 1

Why not directly?:

$$x\equiv y\pmod{p^k}\implies x=y+mp^k\;,\;\;m\in\Bbb Z\implies$$

$$x^p=\sum_{n=0}^p\binom pny^n(mp^k)^{p-n}\;\;(**)$$

But all the binomial coefficients except the first one and the last one are dividisible by $\;p\;$ (either you know this or you can prove it by induction...) , so we get:

$$(**)=(\text{something divisible by}\;p^{k+1})+m^pp^{pk}+y^p=y^p\pmod{p^{k+1}}$$

since, of course, $\;pk>k+1\;$

Answer 2

Suppose $x \equiv y \pmod {p^k}$, for some positive integer $k$.

Note the identity

\begin{align*} x^p - y^p &= (x - y) \left( x^{p-1} + x^{p-2}y + \cdots + xy^{p-2} + y^{p-1} \right)\\[4pt] &=(x - y) \left( \sum_{i=0}^{p-1}x^{p-1-i}y^i \right) \\[4pt] \end{align*}

Then

\begin{align*} &x \equiv y \pmod{p^k}\\[4pt] \implies\; &x \equiv y \pmod{p}\\[4pt] \implies\; &x^i \equiv y^i \pmod{p},\;\;\text{for all nonnegative integers $i$}\\[4pt] \end{align*}

Hence

\begin{align*} \sum_{i=0}^{p-1}x^{p-1-i}y^i &\equiv \sum_{i=0}^{p-1}x^{p-1-i}x^i \pmod{p}\\[4pt] &\equiv \sum_{i=0}^{p-1}x^{p-1} \pmod{p}\\[4pt] &\equiv px^{p-1} \pmod{p}\\[6pt] &\equiv 0 \pmod{p}\\[4pt] \end{align*}

Thus, $\displaystyle{\sum_{i=0}^{p-1}x^{p-1-i}y^i}$ is a multiple of $p$.

But also, by hypothesis, $x - y$ is a multiple of $p^k$, hence

\begin{align*} &p^k\,|\,(x-y)\;\;\;\text{and}\;\;\, p\;{\large{|}}\left(\sum_{i=0}^{p-1}x^{p-1-i}y^i\right)\\[6pt] \implies\; &p^{k+1}\;{\Large{|}} \left((x - y) \left( \sum_{i=0}^{p-1}x^{p-1-i}y^i \right)\right)\\[6pt] \implies\; &x^p - y^p \equiv 0 \pmod {p^{k+1}}\\[6pt] \end{align*}

as was to be shown.

A point of interest: We never used the fact that $p$ is an odd prime, hence the same result holds for any positive integer $p$.