<s>[INST] Find all the integer solutions for: $3x^2+18x+95\equiv 0\pmod {143}$ I need help with the following question:

Find all the integer solutions for: $3x^2+18x+95\equiv 0\pmod {143}$

My solution: First I know that $143=11\cdot 13$ then because $\gcd (11,13) = 1$ then $3x^2+18x+95\equiv 0\pmod {143}$ if, and only if $$3x^2+18x+95\equiv 3x^2+7x+7\equiv 0\pmod {11} \\ 3x^2+18x+95 \equiv 3x^2+5x+4\equiv 0\pmod {13}$$
I don't know how to solve those equations and I don't know how to combine it to the big solution for the real question (I know about the CRT, but I didn't realy understood how to use it, I'd love help with this).
thanks in advance
 [/INST] Let's take one equation
\begin{align*}
3x^2+7x+7 & \equiv 0 \pmod{11}\\
4(3x^2+7x+7) & \equiv 4(0) \pmod{11}\\
x^2+28x+28 & \equiv 0 \pmod{11}&& (\because 4(3) \equiv 1 \pmod{11})\\
x^2+6x+6 & \equiv 0 \pmod{11}&& (\because 28 \equiv 6 \pmod{11})\\
(x+3)^2-3 & \equiv 0 \pmod{11}\\
(x+3)^2-5^2 & \equiv 0 \pmod{11}&& (\because 5^2 \equiv 3 \pmod{11})\\
(x-2)(x+8) & \equiv 0 \pmod{11}
\end{align*}
Since $11$ is prime so if $11 | ab$, then $11$ divides at least one of them, so we get
$$x\equiv 2 \pmod{11} \quad \text{ or } \quad x \equiv -8 \equiv 3\pmod{11}.$$
Likewise (you can work this out yourself)
$$3x^2+5x+4 \equiv 0 \pmod{13} \implies x\equiv 2 \pmod{13} \, \text{ or } \, x \equiv \color{blue}{b}\pmod{13}. $$
So we have the following situation
\begin{align*}
x&\equiv 2 \pmod{11} & x&\equiv 2 \pmod{11} & x&\equiv 3 \pmod{11} & x&\equiv 3 \pmod{11}\\
x&\equiv 2 \pmod{13} & x&\equiv  \color{blue}{b} \pmod{13} & x&\equiv  \color{blue}{b} \pmod{13} & x&\equiv 2 \pmod{13}
\end{align*}
Now use CRT (hopefully you know how to apply it to simple systems like these) to solve these systems.
For example the last system
\begin{align*}
x & \equiv 3 \pmod{11}\\
x & \equiv 2 \pmod{13}
\end{align*}
yields
$$x \equiv 3(13)(6)+2(11)(6) \equiv \color{red}{80} \pmod{143}. $$
Likewise  you will get a total of $\color{red}{4}$ incongruent solutions.
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