If $a \equiv b$ (mod 2n), then $a^2 \equiv b^2$ (mod 4n)

Question

How would I go about proving: If $a \equiv b$ (mod 2n), then $a^2 \equiv b^2$ (mod 4n)?

I already tried proving $a+b = 2nk$ for some integer k, and that was pretty straightforward. But when I try to prove $a-b = 2nk$, I don't know what algebraic trick I need in order to get it to $a^2 - b^2 = 4nz$.

Answer 1

$a-b=2nk$ , where $k$ is an integer. So, $a-b$ is even.

$a=b+2nk$, so $a+b=2b+2nk$ is also even.

Then $a+b=2m$ where $m$ is an integer.

Now $a^2-b^2=2nk\times 2m$. Hence the result.

Answer 2

Hint $\ 2n\mid a-b\,\Rightarrow\,\color{#c00}{2\mid a+b}\,\Rightarrow\,\color{#c00}2(2n)\mid (a-b)(\color{#c00}{a+b}) = a^2-b^2$

Answer 3

Hint: For some $k \in \mathbb Z$, we know $a = b + 2nk$. Compute $a^2$.