<s>[INST] 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. 
 [/INST] $p$ is odd then $x$ and $y$ have different parity. We can assume without loss of generality that $x$ is odd and $y$ is even.
We set $$x=2X+1$$
$$y=2Y$$
Then $$x^2+y^2=4X^2+4X+1+4Y^2=4(X^2+X+Y^2)+1=p$$
Hence, We only have solutions if only:
$$p\equiv 1\pmod 4$$
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