# Is $x^2 ≡ 295$ (mod 2717) solvable?

Is $x^2 ≡ 295$ (mod 2717) solvable?

-Having a tough time with this problem, currently covering a section on Quadratic Reciprocity Law of Gauss. After coming back to my professor, his hint was to consider 2717 ≡ 11*13*19. This wasn't much help unfortunately, any help is really appreciated.

Solving for $x^2\equiv 295\pmod{2717}$, you can instead solve the system for $\begin{cases}x^2\equiv 295\pmod{11}\\x^2\equiv 295\pmod{13}\\x^2\equiv 295\pmod{19}\end{cases}$

Simplifying:

$\begin{cases}x^2\equiv 9\pmod{11}\\x^2\equiv 9\pmod{13}\\x^2\equiv 10\pmod{19}\end{cases}$

This implies that $x\equiv 3\pmod{11}$ or $x\equiv 8\pmod{11}$

$x\equiv 3\pmod{13}$ or $x\equiv 10\pmod{13}$

What about for modulo $19$?

The only squares modulo 19 are $1,4,5,6,7,9,11,16,17$

$10$ is a quadratic residue only for primes $p\equiv 1,3,9,13,27,31,37,39\pmod{40}$
For each choice of outcome, one may apply the Chinese-Remainder Theorem to find what $x$ is modulo $2717$ uniquely. The results will be different depending on which triple is selected.
But in this case, there is no result for what $x$ can be modulo $19$, so $x$ does not exist.