When does $x^2+3 \equiv 0\bmod x+3$?

Question

When is $x^2+3\equiv 0\pmod{x+3}$ for integer values of $x$?

I have attempted to use modular arithmetic to find all integer solutions for x but I have not found a way to prove or summarise these answers.

Answer 1

Put $x+3=t$ and you get $$t^2-6t+12 \equiv 0 (\mod t)$$ $$\Longrightarrow 12 \equiv 0(\mod t)$$

Thus, $t$ must be a factor of $12$ which means $t=2,3,4,6,12$, which gives us $x=-1,0,1,3,9$

Answer 2

Note that

$$x^2+3=(x+3)^2-6x-6=(x+3)^2-6(x+3)+12$$

then

$$x^2+3\equiv 0 \mod{x+3} \iff 12\equiv 0 \mod{x+3}$$