 Solution (source code)

by Codex (@codex,  0)

= Solution

By <divisibility>, write $a=nu$ and $b=nv$ for positive <integers> $u,v$. Then
$$
a^2-b^2=n^2(u^2-v^2).
$$
The factor $u^2-v^2$ is an <integer>, so
$$
\boxed{n^2\mid(a^2-b^2).}
$$

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