<s>[INST] Solving non-linear congruence $x^2+2x+2\equiv{0}\mod(5)$, $7x\equiv{3}\mod(11)$

My attempt:
$x^2+2x+2\equiv{0}\mod(5)$
$(x+1)^2\equiv-1\mod(5)$, we have $x+1\equiv-1\mod(5)$
since $5$ and $11$ are coprime. We have a solution in $\mathbb{Z}_{11}$
With $[3]$ represent $3$, $[13]$ works for $7x\equiv3\mod11$
so $[3]$ is the solution in $\mathbb{Z}_{55}$
General solution is $x=-2+k55$.
But the answer is wrong..
 [/INST] As you said, $(x+1)² \equiv -1 \equiv 4 \pmod 5$, hence $(x+1)\equiv 2 or -2 \pmod 5$. Now, if $(x+1)\equiv -2 \equiv 3 \pmod 5$, then, since also $x\equiv 2 \pmod {11}$ is a solution to the last congruence, one solution is $x\equiv 2 \pmod {55}$. On the other hand, for $x\equiv 1 \pmod {5}$, we should solve for $x\equiv 2 \pmod {11}$. Using Chinese Reminder Theorem, we conclude that the other solution is given by $x\equiv 11-20\equiv -9 \pmod{55}$.
P.S. Here we might view CRT as a decomposition of rings. But in essence, it talks about finding integers $x_i$ such that $\Sigma _ia_ix_i \pmod b$ is a solution of the congruences $x\equiv a_i \pmod {b_i}$, where $b=\Pi_ib_i$.
Reference
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