# A prime congruent to 3 modulo 4 & sums of squares

Prove:

If $p$ is a prime where $p \equiv 3 \pmod{4}$ then $p$ can't be written as the sum of two numbers squared.

I attempted by contradiction, supposing that $p=a^2 + b^2$ where $a,b$ are integers. From definitions i got down to $4$ must divide $a^2 + b^2 - 3$ (since $p=a^2 + b^2\equiv 3 \pmod{4} )$ but I am unsure why this is (or leads to) a contradiction.

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We have that $p\equiv 3 \mod 4$, thus $p=4k+3$. Now, modulo $4$ we have that the squares are $$0^2=0\\1^2=1\\2^2=0\\3^2=1$$
But then $$a^2+b^2=0,1,2\mod 4$$
That is, any sum of squares cannot be $\equiv 3\mod 4$. Note that $p$ being prime is irrelevant!