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

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, and I'm wondering if this will make the proof easier or not.

2 Answers

$$a=b+kp$$ 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 '18 at 23:45

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/