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Problem

Show that if $a equiv bpmod{2n}$, then $a^2equiv b^2pmod{4n}$. More generally, show that if $a equiv bpmod{kn}$, then $a^kequiv b^kpmod{k^2n}$. [Introduction to Higher Arithmetic, ex. 2.1]

Proof of the first statement

Note that $2n mid a - bimplies 2n (a + b)mid a^2 - b^2$ but the $pmod{2n}$ congruence implies that $a$ and $b$ are of the same parity and so $2mid (a + b)$ which implies that $2n cdot 2mid 2n (a + b)mid a^2 - b^2$ or $a^2 = b^2pmod{4n}$.

Attempt at proof of the general case

Multiplying $a equiv bpmod{kn}$ by itself $k$ times, we get $a^k equiv b^kpmod{kn}$, but it is still needed to prove that $k$ divides $a^k - b^k$. It maybe stands reasonable to consider the following expansion: $$(a - b) ^ k = a^k - {k choose 1} a^{k-1}b^1 + {k choose 2} a^{k-2}b^2 - cdots mp {k choose 1} a^1b^{k-1} pm b^k$$ Under $pmod{kn}$, we can replace any $b$ with $a$ and vice versa, which might be useful. Maybe the trick is to sum binomial coefficients to a number (or two numbers) which are divisible by $k$. But sum over all of coefficients is $2^k$, which is not necessarily divisible by $k$.

So, I got stuck here.

Note that $aequiv bpmod{kn}implies a=b+kln$ for some integer $l$. Then $$a^k-b^k=(b+kln)^k-b^k=binom k1b^{k-1}kln+binom k2b^{k-2}(kln)^2+cdots+(kln)^k$$ Every term has a factor of $k^2n$ from $(kln)^t$ for $1<tle k$ but $dbinom k1=k$ so $k^2nmid (a^k-b^k)$.