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I'm not really sure how to approach this problem, since it doesn't seem similar to solving linear congruences in $\mathbb{Z}_m$.

Find all solutions in $\mathbb{Z}_5[x]$ to the congruence $(x^2-1)a(x)\equiv x^2+x-2\pmod{x^3-1}$. Additionally, is it possible to count the number of solutions in $(\mathbb{Z}_5[x])_{x^3-1}$ without actually finding them?

Any help is appreciated.

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Actually, it is similar to solving linear congruences in $\mathbb Z/(n)$. In this case, you are starting with the principal ideal domain (PID) $\mathbb Z$, and expressing $Z/(n)$ as the sum of the prime-power factors, as $\mathbb Z/(72)\cong\mathbb Z/(8)\,\oplus\,\mathbb Z/(9)$, that is, you find solutions good modulo $8$ and $9$ separately and use Chinese Remainder Theorem. Right?

In your case I’ll use $\mathbb F_5$ to denote the field with five elements. It’s the same for the PID $\mathbb F_5[x]$ as it was for $\mathbb Z$. You have an element of the ring, namely $x^3-1$, that factors as $(x-1)(x^2+x+1)$, and you’ll need to solve your linear congruence modulo $x-1$ and modulo $x^2+x+1$. It’s messier, I’m sure, but the principle is exactly the same as in the familiar case.

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  • $\begingroup$ Thanks, but I'm not really sure how to go about solving the actual congruences. $\endgroup$ – Javaris JL. Nov 3 '13 at 2:48
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    $\begingroup$ @JavarisJL, the congruence modulo $x-1$ is easy, because modulo this polynomial, $x$ is as good as $1$, so that the equation becomes $0\cdot a=0$, thus valid for all $a$. The second one is a little trickier; you see that modulo $x^2+x+1$, $x^2-1$ is congruent to $4x+3$, and you need the reciprocal of this to solve your congruence. I do leave that to you. $\endgroup$ – Lubin Nov 3 '13 at 3:19

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