How can I solve $2x+\sqrt {10} \equiv 10 \pmod{13}?$

Question

I have seen such problem:

Find the value of $x$: $$2x+\sqrt {10} \equiv 10 \pmod{13}.$$

I think, this problem is wrong, isn't it?

$$\frac {2x+\sqrt{10}-10}{13} \in\mathbb{Z^{+}}.$$

I do not understood, what will I do with $\sqrt{10}$?

Answer 1

The definition of $\sqrt{10}\bmod 13$ is an integer $x$ with $x^2\equiv 10\bmod 13$, or $x^2-10=0$ in the finite field $\mathbb{F}_{13}$. This polynomial has $2$ solutions in this field, namely, $$ x^2-10=(x+6)(x+7)=0. $$ So there are two solutions for $\sqrt{10}$.