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Given $x^2 - \overline{51}x - \overline{43} = \overline{0}$.

Solve it in $\mathbb{Z}/187\mathbb{Z}$.

First of all, does the $\overline{x}$ mean that $\overline x = \{x + z \ | z\in I\}$?

I am getting really confused, because 187 is composite. I think I should somehow decompose this equation into two equations and solve them in $\mathbb{Z}/11\mathbb{Z}$ and $\mathbb{Z}/17\mathbb{Z}$, but I'm not sure.

And what is the pattern to solve things like that?

Bill Dubuque
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False Promise
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    Yes, when working with a modulus with more than one prime factor, you may always decompose the congruence into separate congruences modulo primes (or prime powers), and solve them independently. Then you reconstitute the solutions by Chinese Remainder Theorem. Note that if each congruence has two roots, then there are a total of 4 ways to reconstitute them. – Erick Wong Dec 10 '16 at 02:10

2 Answers2

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Yes, find the roots $\,r_1,r_2\,$ mod $17,\,$ and the roots $\,s_1,s_2\,$ mod $11$. Then by CRT (see Remark below) each of the four pairs $\,(r_i,s_j)\,$ corresponds to a unique root mod $\,17\cdot 11.\,$

Easiest is to solve the CRT system generically. Suppose $\,x\equiv r\pmod{\!17}\,$ and $\,x\equiv s\pmod{\!11}.\,$ Then $\, x\equiv r\!+\!17j,\,$ so ${\rm mod}\ 11\!:\ s\equiv x\equiv r\!+\!17j\equiv r\!+\!6j\!\iff\! 6j\equiv s\!-\!r\!\iff\! \color{#c00}{j\equiv 2(s\!-\!r)}.\,$ Thus $\, x=r\!+\!17\color{#c00}j = r+17(\color{#c00}{2(s\!-\!r)}+11k) = \color{#0a0}{34s-33r} + 17\cdot11\,k$

Finally, $ $ plug all four values of $\,r_i,\,s_j\,$ into $\, x=\color{#0a0}{34s-33r}\,$ to get all four roots mod $\,17\cdot11$.

You'll end up discovering $\ \ \bbox[8px,border:2px solid blue]{ f(x) \equiv (x\!-\!3)(x\!-\!48)\equiv (x\!-\!14)(x\!-\!37) \ \ \pmod{187}}$

Remark $\ $ 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.

$$\begin{eqnarray} f(x)\equiv 0\pmod{mn}&\overset{\rm CRT}\iff& \begin{array}{}f(x)\equiv 0\pmod m\\f(x)\equiv 0\pmod n\end{array} \\ &\iff& \begin{array}{}x\equiv r_1,\ldots,r_k\pmod m\phantom{I^{I^{I^I}}}\\x\equiv s_1,\ldots,s_\ell\pmod n\end{array}\\ &\iff& \left\{ \begin{array}{}x\equiv r_i\pmod m\\x\equiv s_j\pmod n\end{array} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}^{\phantom{I^{I^{I^I}}}}\\ &\overset{\rm CRT}\iff& \left\{ x\equiv t_{i j}\!\!\pmod{mn} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}\\ \end{eqnarray}\qquad\qquad$$

Bill Dubuque
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HINT.-$x^2-51x-43=0$ modulo $187=11\cdot17$ You have $$\begin{cases}x^2+4x+1=(x-3)(x-4)=0 \text{ in } \Bbb F_{11}\\x^2+8=(x-3)(x-14)=0 \text{ in } \Bbb F_{17}\end{cases}$$ Try now to apply chinese theorem.

Piquito
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