Show that for $p$ to be odd prime and $p \equiv 3$ mod $4$, then $x^2+y^2 = p$ has no integer solution Show that for $p$ to be odd prime and  $p \equiv 3$ mod $4$, then $x^2+y^2 = p$ has no integer solution. I have no idea how can i apply quadratic reciprocity to the equation  $x^2+y^2 = p$ or should use other method. 

Generalization,
If $k$ be a positive integer satisfying $k \equiv 3 \pmod 4$ then $x^2+y^2=k$ has no integer solutions.
Proof
$k \equiv 3 \pmod 4$ is given and if $x,y$ be integers satisfying $x^2+y^2=k$ then we must have, 
$x^2+y^2\equiv 3 \pmod 4$
But,
Given any integer $j$ we know that $j^2 \equiv 0 \pmod 4$ or $j^2 \equiv 1 \pmod 4$
And hence $x \in \mathbb{Z} , y \in \mathbb{Z}$ implies, $x^2+y^2 \equiv 0+0 = 0 \pmod 4$ or $\equiv 1+0 = 1 \pmod 4$ or $\equiv 1+1 = 2 \pmod 4$or $\equiv 0+1 = 1 \pmod 4$ 
And again, None of the numbers $0,1,2$ is $\equiv 3 \pmod 4$ .
Hence contradiction !!
Thus our assumption was false.
Hence Proved
