$a \equiv b$ (mod $p$) implies $a^{p^n} \equiv b^{p^n}$ (mod $p^n$)?

Question

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.

Answer 1

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

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

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

Answer 2

There is following theorem (Lifting The Exponent Lemma). We will use notation $p^{\alpha}||n$ for positive integer $n$, nonnegative integer $\alpha$ and prime $p$, which equivalent to $p^{\alpha}|n$ and $p^{\alpha+1}\nmid n$.

Theorem. Let $a,b,n$ be a positive integers and $p^{\alpha}||a-b$, $p^{\beta}||n$. Then:

• if $p>2$ and $\alpha\geq 1$ then $p^{\alpha+\beta}||a^n-b^n$;
• if $p=2$ and $\alpha\geq 2$ then $2^{\alpha+\beta}||a^n-b^n$.

Note. If $p=2$ and $\alpha=1$ then $2^{\beta+1}||a^n-b^n$.

Your statement is a consequence of this theorem.

Some links about LTE lemma:

What can I do with the lifting the exponent lemma?

https://brilliant.org/wiki/lifting-the-exponent/