For which n does a quadratic congruence have solutions e.g $x^2=83 \pmod n$
I've tried using the Chinese remainder theorem to break it down, this is what I have so far
$x^2=83 \pmod2 $ has solutions
$x^2=83 \pmod4 $ has no solutions
$x^2=83 \pmod{83}$ has solution
$x^2=83 \pmod{83^2}$ has no solutions
suppose $p\neq 2,83$ then use quadratic reciprocity
$(83/p)=(-1)^{(83-1)(p-1)/4} (p/83)$
Not sure what to do next