# Questions tagged [polynomial-congruences]

Questions about congruences where the modulus is a polynomial. For questions concerning congruences between polynomials where the modulus is an integer, use the tag (modular-arithmetic) instead.

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### Most efficient solution to find polynomial congruence for 0 mod p

I was given the polynomial $$f(x) = x^4 + 2x^3 + 3x^2 + x + 1$$ and told to find $$f(x) \mod 17 = 0$$ I found the solution to be $$x = 8 + 17n$$ However, I arrived at this solution by computing all ...
1answer
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### Let $f(x)=x^3+x^2-5$. Show that for $n=1, 2,3, …$ there is a unique $x_n$ modulo $7^n$ such that $f(x_n)\equiv 0\pmod{7^n}$.

My gut feeling for solving this problem is to use strong induction. Starting with the base case $n=1$ we can check each of the seven congruence classes and find that $x_1=2$ is the unique solution. ...
1answer
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### How to solve a non-linear system of modulus equations?

I have the following problem: $$2x^2 + 8 \equiv 6 \;(\bmod\;13)$$ $$x \equiv 2 \;(\bmod\;15)$$ I have tried applying the Chinese remainder theorem, but could not figure out how to make it work, as ...
1answer
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2answers
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### solve $x^2 -4x +13 \equiv 0 \pmod{81}$?

How do I solve $x^2 -4x +13 \equiv o \pmod{81}$ ? I know that this is the same as $x^2 -4x +13 \equiv x^2 + 2x + 1 \equiv (x +1)^2\equiv 0\pmod{3^4}$ but why is $x \equiv -1\pmod{3}$ the only ...
2answers
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### Is there a solution for $x^6 \equiv 5 \pmod {71}$? [closed]

How can I verify that a solution exists for $$x^6 \equiv 5 \pmod {71}$$
1answer
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### What is the congruence class of $x^3\mod x^3+x+1$?

I have a given Polynom congruence with a Polynom $x^3+x+1$ ... so the set of the congruence classes is $\{0, 1,x,x+1,x^2,x^2+1,x^2+x,x^2+x+1\}$ But what would look this like? x^3\mod x^3+x+1\equiv ...
0answers
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### Need hints on the following algebra problems.

I've been looking at these for over an hour and I don't understand how to do them. Any hints would be greatly appreciated. Let $p(x) = x^3 + x + 1$ and $F = Z_3[x]/\langle p(x)\rangle$. Factor \$p(x)...