Solve $x^2+5x+6 \equiv 0 \pmod{\!11\cdot 17}$ Solve $x^2+5x+6 \equiv 187 \mod 187$
Solution
$$x^2+5x+6 \equiv 187 \mod 187$$
$$ (x+\frac{5}{2})^2 \equiv \frac{1}{4}$$
$$ 4(x+\frac{5}{2})^2 \equiv 1$$
$$ y:= x+\frac{5}{2} $$
$$ 4y^2 \equiv 1 \mod 11 \wedge 4y^2 \equiv 1 \mod 17  $$
$$ ( 2y \equiv 1 \mod 11 \vee 2y \equiv 10 \mod 11 ) \wedge ( 2y \equiv 1 \mod 17  \vee 2y \equiv 13 \mod 17)  $$
$$ ( y \equiv 6 \mod 11 \vee y \equiv 5 \mod 11 ) \wedge ( y \equiv 9 \mod 17  \vee y \equiv 15 \mod 17)  $$
Combining that from CRT I got:
$$ y \in \left\{49, 60,83,94 \right\} $$
and for example:
$$ x+\frac{5}{2}  \equiv 94 \mod 187$$
$$ 2x  \equiv 183 \mod 187$$
some calculus and get...
$$x \equiv 185 $$
And the same thing for each other case.
Question
Is there any faster (or smarter) way to solve equations like that?
 A: Since $x^2+5x+6=(x+2)(x+3)$,\begin{align}x^2+5x+6\equiv187\pmod{187}&\iff(x+2)(x+3)\equiv0\pmod{187}\\&\iff(11\mid x+2\vee11\mid x+3)\wedge(17\mid x+2\vee17\mid x+3).\end{align}
A: Hint: applying CRT as in the Remark below yields
$$(x\!+\!2)(x\!+\!3)\equiv 0\!\!\!\pmod{\!11\cdot 17}\iff \begin{align} x\equiv -2,-3\!\!\!\pmod{\!11}\\ x\equiv -2,-3\!\!\!\pmod{\!17}\end{align}\qquad\qquad $$
which combine to $4$ solutions $\,x\equiv (\color{#90f}{{ -2,-2}}),\,(\color{#0a0}{-3,-3}),\,(-2,-3),\,(-3,-2)\,$ mod $(11,17).\,$  By CCRT, $\bmod 187\:$  the first two yield $\,x\equiv \color{#90f}{ -2}\,$ and $\,\color{#0a0}{-3}.\,$ The third case  $\,(-2,-3)\,$ yields by CRT:
$\!\bmod\, \color{#c00}{11}\!:\,\  {-}2 \equiv\, x \equiv -3+17\,\color{#c00}k \equiv-3+6k \iff 6k\equiv1\equiv12 \iff \color{#c00}{k \equiv 2}$
therefore we infer $\ x = -3+17(\color{#c00}{2+11}n) = 31+187n,\ $ so $\ (-2,-3)\,\mapsto 31$
Finally   $\ (-3,-2) + \underbrace{(-2,-3)}_{\large31}\equiv \underbrace{(-5,-5)}_{\large -5}$ $\,\Rightarrow\,(-3,-2)\,\mapsto\, -5-31\equiv -36\ $
Remark $ $ Note that $\,x\!+\!3\,,x\!+\!2\,$ are coprime (having difference $=1),\,$ so a prime power $p^n$ divides their product iff it divides exactly one of them, so the above method still works if we replace $11$ and $17$ by powers of distinct primes.
Generally, if $\,m,n\,$ are coprime then, by CRT, solving a polynomial $\,f(x)\equiv 0\pmod{\!mn}\,$ is equivalent to solving $\,f(x)\equiv 0\,$ mod $\,m\,$ and mod $\,n.\,$ By CRT, each combination of a root $\,r_i\bmod m\,$ and a root $\,s_j\bmod n\,$ corresponds to a unique root $\,t_{ij}\bmod mn,\,$ i.e.
$$\begin{eqnarray} f(x)\equiv 0\!\!\!\pmod{\!mn}&\overset{\,\,\rm CRT}\iff& \begin{array}{}f(x)\equiv 0\pmod{\! m}\\f(x)\equiv 0\pmod{\! n}\end{array} \\ 
&\,\,\iff&  \begin{array}{}x\equiv r_1,\ldots,r_k\pmod{\! m}\phantom{I^{I^{I^I}}}\\x\equiv s_1,\ldots,s_\ell\pmod{\! n}\end{array}\\ 
&\,\,\iff& \left\{ \begin{array}{}x\equiv r_i\pmod{\! m}\\x\equiv s_j\pmod {\! n}\end{array} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}^{\phantom{I^{I^{I^I}}}}\\
&\overset{\,\,\rm CRT}\iff& \left\{ x\equiv t_{i j}\!\!\pmod{\!mn} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}\\
\end{eqnarray}\qquad\qquad$$
For more complex examples it is usually easier to solve the CRT system first for generic (symbolic) roots, then plug in the specific root values for all combinations, e.g. see here and here.
A: $$x^2+5x+6\equiv187\equiv0 \pmod {187=11\times17}$$
$$(x+2)(x+3)\equiv 0 \pmod {11  , 17}$$
$$x\equiv-2 \text { or } -3 \pmod {11, 17}$$
Now use the Chinese Remainder Theorem.
A: Solve $x^2+5x+6 \equiv (x+2)(x+3) \pmod {187}$.
Besides the two 'in your face' solutions, $x \equiv -2 \pmod{187}$ and $x \equiv -3 \pmod{187}$, we can buttress the argument given by José Carlos Santos to find all four solutions.
We want to find integers $x, k, j$ satifying
$\;\text{L1:}\quad x + 2 = 11k$
$\;\text{L2:}\quad x + 3 = 17j$
Subtracting $\text{L1}$ from $\text{L2}$ we write
$\tag 1 1  = 17j - 11k$
Bézout's identity gives us
$\quad 1 = 17(2) - 11(3)$
So $k = 3$ and, plugging into $\text{L1}$,  $x \equiv 31 \pmod{187}$ is a solution.
For the last solution,
$\;\text{L1:}\quad x + 2 = 17j$
$\;\text{L2:}\quad x + 3 = 11k$
Subtracting $\text{L1}$ from $\text{L2}$ we write
$\tag 2 1  = -17j + 11k$
Bézout's identity gives us
$\quad 1 = 17(2) - 11(3)$
So $k = -3$ and, plugging into $\text{L2}$,  $x = -36 \equiv 151 \pmod{187}$  is a solution.
It can also be argued that there are exactly four solutions.
