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I'm finding the solutions for the modular equation $x^2 -3x+2 \equiv 0\pmod{14}$. This is what I've done so far:

\begin{align} 0 & \equiv x^2 -3x+2 \pmod{14}\\ & \equiv (x-1)(x-2) \pmod{14} \\ \end{align} This implies that two of the solutions are the polynomial's usual roots: 1, 2.

This is where I get lost. I realize I should be using the following fact to find the other solutions, but don't know how to approach it.$$(x-1)(x-2) \overset{14}{\equiv} 0 \Leftrightarrow 14|(x-1)(x-2)$$ The provided answer says that 8,9 are the other solutions. How would I go about finding these?

J. W. Tanner
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Josh Katofsky
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    Solve $2|x-1$ and $7|x-2$ and vice versa – J. W. Tanner Mar 03 '20 at 21:47
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    Where do those come from? Because of $2 \times 7 = 14$? @J.W.Tanner – Josh Katofsky Mar 03 '20 at 21:52
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    In general you can approach this type of congruence equations with composite modulus by solving first $0\equiv x^2-3x+2\pmod{2}$ and $0\equiv x^2-3x+2\pmod{7}$. You would get solutions of the form $x$ satisfying $x\equiv a\pmod{2}$ and $x\equiv b\pmod{7}$. With a prime modulus an equation like $0\equiv(x-1)(x-2)\pmod{p}$ implies that there are only the $2$ solutions that you expect. Then you can use the Chinese Remainder Theorem to combine these two conditions to the form $x\equiv c\pmod{14}$. –  Mar 03 '20 at 21:55

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If $14|(x-1)(x-2)$, we cannot conclude $14|x-1$ or $14|x-2$, because $14$ is not prime.

But $14=2\times7$, so $14|(x-1)(x-2)$ means $2|(x-1)(x-2)$ and $7|(x-1)(x-2)$.

Since $2$ and $7$ are prime, this means ($2|x-1 $ or $2|x-2$) and ($7|x-1$ or $7|x-2$).

Thus, there are four solutions (mod $14$). Can you take it from here?

J. W. Tanner
  • 60,406