I'm trying to solve the following problem.
Find the solutions of
$x^2 + 2x +2 = 0$ in the ring $\mathbb{Z}/1105\mathbb{Z}$.
I don't know how to solve this. I surely can't check all the possibilities.
Ok. I've now split it up. and found:
$(x+1)^2 = 4 \bmod 5, \quad x = 1,2 \bmod 5$
$(x+1)^2 = 12 \bmod 13, \quad x =4,7 \bmod 13$
$(x+1)^2 = 16 \bmod 17, \quad x= 3, 12 \bmod 17$
but I don't know how to get the answer using the CRT
Hint: there's a formula for the solutions of a system of two linear congruences with coprime moduli $m$ and $n$: $$\begin{cases}x\equiv a \pmod m, \\ x\equiv b \pmod n.\end{cases} $$ Start from a Bézout's identity $\; um+vn=1$. The solutions are given by this formula: $$x \equiv bum+avn\pmod{mn}.$$ So you can first obtain the solutions mod. $5\times 13$, and thence the solutions mod. $65\cdot 17$. In all you'll have $8$ roots for the quadratic equation.