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I'm struggling to understand how the Chinese remainder theorem works on quadratics (though I know how it works in the simpler cases). For example in my number theory book there's an example with $f(x) = x^{2} + x + 7$ and we need to find all the solutions to $f(x) \equiv 0$ (mod $189$) given that the roots mod $27$ are $4, 13, 22$ and the roots mod $7$ are $0$ and $6$.

There's a solution in the book to the example but I can't understand it very well. It says: "...we find that $x \equiv a_{1}$ (mod $27$) and $x \equiv a_{2}$ mod($7$) if and only if $x \equiv 28 \cdot a_{1} - 27 \cdot a_{2}$ (mod $189$)."

And then the solution (there are six) follows easily by plugging in the solutions provided in the mod $27$ and mod $7$ cases listed above.

But my main struggle is understanding how this equation $x \equiv 28 \cdot a_{1} - 27 \cdot a_{2}$ (mod $189$) was derived, and how to get this equation in general for these problems (I think it involves Chinese remainder theorem and Euclidean algorithm but I can't figure it out).

Wilson Brians
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  • The complete solution of an analogous example is in the linked duplicate. If you have further questions please post comments there (or here). – Bill Dubuque Oct 21 '18 at 14:34

1 Answers1

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This comes from the reverse isomorphism in the Chinese remainder theorem: if $a$ and $b$ are coprime and $ua+vb=1$ is a Bézout's relation between $a$ and $b$, this reverse isomorphism is given by \begin{align} \mathbf Z/ab\mathbf Z &\longleftarrow \mathbf Z/a\mathbf Z \times \mathbf Z/b\mathbf Z \\ ua\beta+vb\alpha \bmod ab&\longleftarrow\mkern-8mu|\,(\alpha\bmod a,\beta\bmod b) \end{align}

Bernard
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