Solution of $2x^2 \equiv -21 \space \pmod {79}$ in $\mathbb{Z}_{79}$

Question

I need to find whether this equation has a solution in $\mathbb{Z}_{79}$ (p-adic integers) - $$2x^2 \equiv -21 \space \pmod {79}$$

Now if instead of $2x^2$ I had $x^2$ then I simply need to check whether $-21$ is a quadratic residue $mod \space 79$. But here what should I do? Any suggestions!

Answer 1

$$2x^2\equiv -21\equiv -100\pmod{79}\implies x^2 \equiv -50\pmod{79}$$ $$-50 \equiv -2\times 25 \equiv -1\times 81\times 25$$ As $79\equiv 3\pmod4$, $-1$ is not a quadratic residue in $\bmod{79}$, thus, there is no such $x$.