Difference of Two Squares cannot equal 2 modulo 4/Proof 1

Theorem

Let $n \in \Z_{>0}$ be of the form $4 k + 2$ for some $k \in \Z$.

Then $n$ cannot be expressed in the form:

$n = a^2 - b^2$

for $a, b \in \Z$.

Proof

Let $n = a^2 - b^2$ for some $a, b \in \Z$.

By Square Modulo 4, both $a$ and $b$ are of the form $4 k$ or $4 k + 1$ for some integer $k$.

There are $4$ cases:

$a \equiv b \equiv 0 \pmod 4$

Then:

$a^2 - b^2 \equiv 0 \pmod 4$

and so $n$ is in the form $4 k$.

$a \equiv 0 \pmod 4$, $b \equiv 1 \pmod 4$

Then:

$a^2 - b^2 \equiv -1 \pmod 4 \equiv 3 \pmod 4$

and so $n$ is in the form $4 k + 3$.

$a \equiv 1 \pmod 4$, $b \equiv 0 \pmod 4$

Then:

$a^2 - b^2 \equiv 1 \pmod 4$

and so $n$ is in the form $4 k + 1$.

$a \equiv b \equiv 1 \pmod 4$

Then:

$a^2 - b^2 \equiv 0 \pmod 4$

and so $n$ is in the form $4 k$.

Thus it is never the case that $a^2 - b^2 = 4 k + 2$.

$\blacksquare$