The number of incongruent solutions of $x^4-14x^2+36\equiv 0\pmod{2019}$ I was asked to find the number of the incongruent solutions of $x^4-14x^2+36\equiv 0\pmod{2019}$, and the fact $2019=673\times 3$ is given.
My attempt: Given congruence is equivalent to
$$
\begin{cases}
x^4-14x^2+36\equiv 0\pmod{673}\\
x^4-14x^2+36\equiv 0\pmod{3}
\end{cases}
$$
The second one is easily reduced to $1-14\equiv 0\pmod{3}$ if $3\nmid x$, so $x\equiv 0\pmod{3}$. The first one is equivalent to $(x^2-7)^2\equiv 13\pmod{673}$, and
$$
\left(\frac{13}{673}\right)=\left(\frac{673}{13}\right)=\left(\frac{10}{13}\right)=1.
$$
Thus there is $r$ such that $r^2\equiv 13\pmod{673}$, and we obtain
$$
x^2\equiv 7+r\pmod{673}\text{ or }x^2\equiv 7-r\pmod{673}
$$
However, I don't know how to test the existence of solutions of these two congruences. How to do it, or are there other ways?

I found a solution without finding the square root of 13 modulo 673 directly. $x^4-14x^2+36\equiv 0\pmod{673}$ is equivalent to
$$
x^4-12x^2+36\equiv 2x^2\pmod{673}.
$$
Also, there is an integer $r$ such that $r^2\equiv 2\pmod{673}$, since $\left(\dfrac{2}{673}\right)=1$. Thus given congruence is factored to
$$
(x^2-rx-6)(x^2+rx-6)\equiv 0\pmod{673}.
$$
Since 673 is prime, $x^2+rx-6\equiv 0\pmod{673}$ or $x^2-rx-6\equiv 0\pmod{673}$. Let $y=2x\pm r$, then each congruence is equivalent to $y^2\equiv r^2+24\equiv 26\pmod{673}$. Since
$$
\left(\dfrac{26}{673}\right)=\left(\dfrac{2}{673}\right)\left(\dfrac{13}{673}\right)=1,
$$
each congruence has two solutions. Therefore $x^4-14x^2+36\equiv 0\pmod{2019}$ has four incongruent solutions.
