Let $p$ be a prime number. If $a \equiv b$ (mod $p$), does that imply $a^{p^n} \equiv b^{p^n}$ (mod $p^n$)?

I think the answer will be yes, and I suspect that the way of proving it will involve writing $a^{p^n}-b^{p^n}$ as a multiple of $(a-b)^n$.

I also noticed that $n<p^n$, and I'm wondering if this will make the proof easier or not.

up vote 2 down vote accepted

$a=b+kp$ where $k$ is an integer

$(b+kp)^{p^n}=b^{p^n}+\binom{p^n}1b^{p^n-1}kp+\binom{p^n}2b^{p^n-2}(kp)^{2}+\cdots+(kp)^{p^n}$

$\equiv b^{p^n}\pmod{p^{n+1}}$

  • So simple yet so useful! Thank you! – Pascal's Wager Nov 8 at 23:45

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