Note that $2717 = 11*13*19$ and determine if $x^2 \equiv 295$ (mod $2717$) is solvable.
I know I have to spilt this up into three different congruences $x^2 \equiv 295$ (mod $11$), $x^2 \equiv 295$ (mod $13$), and $x^2 \equiv 295$ (mod $19$). Then I solved $(\frac{295}{11})=1, $$(\frac{295}{13})=1$, and $(\frac{295}{19})= -1$. Hence, $x^2 \equiv 295$ (mod $2717$) is not solvable.