Difference of Two Squares cannot equal 2 modulo 4/Proof 1
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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 $a^2 - b^2 = 4 k + 2$.
$\blacksquare$