Sum of Two Squares not Congruent to 3 modulo 4
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Theorem
Let $n \in \Z$ such that $n = a^2 + b^2$ where $a, b \in \Z$.
Then $n$ is not congruent modulo $4$ to $3$.
Proof
Let $n \equiv 3 \pmod 4$.
Aiming for a contradiction, suppose $n$ can be expressed as the sum of two squares:
- $n = a^2 + b^2$.
From Square Modulo 4, either $a^2 \equiv 0$ or $a^2 \equiv 1 \pmod 4$.
Similarly for $b^2$.
So $a^2 + b^2 \not \equiv 3 \pmod 4$ whatever $a$ and $b$ are.
Thus $n$ cannot be the sum of two squares.
The result follows by Proof by Contradiction.
$\blacksquare$