Suppose for all $n\in\mathbb Z$, we have $(x + 4n)^2\equiv x^2\bmod m$. Find all $m\in\mathbb N$ for which this is a true statement.
I have no idea how to go about finding m. I tried to use the fact that $(x+4n)^2 - x^2$ should be divisible by m, and then used the well defined-ness of $+$ and $×$ operations to deduce something but I failed.
$$\forall n:8nx+16n^2\equiv0\bmod m$$ Since $x$ is a variable, we must have $m\mid8n$ and $m\mid16n^2$. We can ignore the second condition as it is implied by the first one. Since $n$ can be any integer, including 1, we must have $m\mid8$, so $m=1,2,4,8$.