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Problem

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

Proof of the first statement

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

Attempt at proof of the general case

Multiplying $a \equiv b\pmod{kn}$ by itself $k$ times, we get $a^k \equiv b^k\pmod{kn}$, 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.

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marked as duplicate by Bill Dubuque elementary-number-theory Jan 4 at 15:33

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Note that $a\equiv b\pmod{kn}\implies a=b+kln$ 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 $1<t\le k$ but $\dbinom k1=k$ so $k^2n\mid (a^k-b^k)$.

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