Solve $x^2+5x+6 \equiv 0 \pmod{\!11\cdot 17}$

Solution 1:

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}

Solution 2:

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.

Solution 3:

$$x^2+5x+6\equiv187\equiv0 \pmod {187=11\times17}$$

$$(x+2)(x+3)\equiv 0 \pmod {11 , 17}$$

$$x\equiv-2 \text { or } -3 \pmod {11, 17}$$

Now use the Chinese Remainder Theorem.