Critique my proof of: For every integer $x$, the remainder when $x^2$ is divided by $4$ is either $0$ or $1$

Critique my proof on structure, correctness, etc.

Proof. Let x be any arbitrary integer.

Case #1

Suppose $x$ is even. Thus, $x = 2k$ and $x^2 = 4k^2$. The quotient of $x^2$ and $4$ is $k^2$, so the remainder is zero.

Case #2

Suppose $x$ is odd. Thus, $x = 2k + 1$ and $x^2 = 4(k^2 + k) + 1$. The quotient of $x^2$ and $4$ is $k^2 + k$ with a remainder of one.

After exhausting all cases, it is evident that for every integer $x$, the remainder when $x^2$ is divided by $4$ is either $0$ or $1$.

Your proof is completely correct. One way to unify the two cases and thereby make it marginally "slicker" is to use the Division algorithm (which is also a way to guarantee that your two cases covers every eventuality).

For example: by the Division algorithm, we can write $x = 2 k + r$ for some (unique) integers $k$ and $r$ with $r = 0$ or $r = 1$. In either case, $$ x^2 = (2 k + r)^2 = 4k^2 + 4k r + r^2 = 4(k^2 + k r) + r^2, $$ and so the remainder of $x^2$ when divided by $4$ is $r^2 = 0$ or $1$, since $r = 0$ or $1$.

Though there is of course no meaningful difference between this and your proof, other than writing both cases in one go.