<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] Simplest way is to note that if $x = 2k + m; m = 0,1$ and $y= 2j + n; n=0,1$ then $x^2 + y^2 = 4(k^2 + j^2 + km + jn) + m^2 + n^2 \equiv m^2 + n^2 \equiv 0,1,2 \mod 4 \not \equiv 3\mod 4$
$p$ being prime has nothing to do with it.
... or $x \equiv 0, \pm 1, 2 \mod 4 \implies x^2 \equiv 0,1 \mod 4$ so $x^2 + y^2 \equiv 0,1,2 \mod 4$.
Not sure which is easier.
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