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Let's consider the following polynomial over $\mathbb Z_p[x]$:

$$f_p = \overline{20}x^4-x^2+\overline{3}$$

I need to:

  • find the $p$ values so that $p$ is prime and $f_p$ can be divided by $g=x^2-x-\overline{2}$;
  • find $p$ values such that $p$ is prime and $f_p$ can be divided by $h = x \cdot g=x^3-x^2-\overline{2}x$.

My attempt to solution

Part One.

I've divided $f_p$ by $g$ and I get a quotient $q = \overline{20}x^2+\overline{20}x+\overline{59}$ and a remainder $r = \overline{99}x+\overline{121}$. If $f_p$ divides $g$ then its remainder must be $0$. So I need to solve:$$\overline{99}x+\overline{121} = 0 \pmod p$$ in order for that to havea solution then $d = \gcd(\overline{99}, p) \mid 121$ and that happens as $p = 11$.

Part Two

If I try to divide $f_p$ by $h$ then I get a degree 2 conguential equation, which is pretty hard to solve. So I figured that maybe I just need to divide the previous quotient by $x$, which leaves me with the easier equation:

$$20x+59 = 0 \pmod{p}$$

but the problem is $ \gcd(20,p) \mid 59 \Leftrightarrow \gcd(20,p) = 59$ and there isn't such value therefore $\nexists p$.

I have the feeling that something is quite wrong with part two of my solution, but I can't tell why, so I am asking you: am I correct? As of part one: is it also correct? Could I have got the same result in a faster way?

haunted85
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1 Answers1

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Method 1: do divide $f_p$ by $h$ to get the degree-2 remainder. For $h$ to divide $f_p$, every coefficient of that degree-2 remainder must be divisible by $p$ (this is a direct generalization of the previous case). So you can take the greatest common divisor of the three coefficients in the remainder.

Method 2: $h$ divides $f_p$ if and only if both $x$ and $g$ divide $f_p$ (this is because $\mathbb Z_p[x]$ is a unique factorization domain). You've already worked out the prime $p$ for which $g$ divides $f_p$. For which primes does $x$ divide $f_p$?

Greg Martin
  • 78,820