# Solving simple diophantine equation with modulos

I need to solve this diophantine equation using a positive integer $x$: $$x^2 + 42x + 21 \equiv 0 \mod 105$$

I think it will be easier if I could use the prime factors of $105$ to get a system of congruences and then use the Chinese Remainder Theorem. Can I do that? If so, why exactly?

If it's possible, then I would get the factors $3$, $5$ and $7$. But I'm getting stuck in this one: $$x^2 + 42x + 21 \equiv 0 \mod 5$$

Can anyone show me how to do it properly?

$x^2+42x+21\equiv x^2+2x+1\pmod 5$
$x^2+42x+21\equiv x^2\pmod{3,7}$