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How would one do this? I know since it doesn't have roots it can only be divisible by irreducible polynomials of degree 2. How would I prove that it is or it isn't? There are a lot of polynomials with this conditions.

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    $\begingroup$ Just try to solve for the coefficients. if it factored, we'd have $p(x)=(x^2+ax+b)\times (x^2+cx+d)$ $\endgroup$
    – lulu
    Jan 2, 2019 at 22:12

2 Answers 2

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The product of all the monic, irreducible polynomials over $\mathbb{F}_7$ wih degree $\leq 2$ is given by $x^{49}-x$, so if we prove that $f(x)=(x^2+1)^2-x$ and $x^{48}-1$ are coprime over $\mathbb{F}_7[x]$ we are done. Let us compute $x^{48}-1\pmod{f(x)}$, for instance through the Brauer chain $x^4\to x^8\to x^{16}\to x^{32}\to x^{48}$. $$ x^4 \equiv -2x^2+x-1\pmod{f(x)} $$ $$ x^8 \equiv 3x^3-3x^2+2x-3\pmod{f(x)} $$ $$ x^{16} \equiv x^3+3x^2+2x-1\pmod{f(x)} $$ $$ x^{32} \equiv -x^3+2x^2+x-3\pmod{f(x)} $$ $$ x^{48}-1 \equiv -x^3-2x^2+x+2 \pmod{f(x)}$$ give $$\gcd(x^{48}-1,x^4+2x^2-x+1)=\gcd(x^3+2x^2-x-2,x^4+2x^2-x+1) $$ $$ = \gcd(x^3+2x^2-x-2,-x-3) $$ but $x=4$ is not a root of $x^3+2x^2-x-2$ since $7\nmid 90$ and we have finished.

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  • $\begingroup$ See this answer for a generalization - an efficient algorithm often used in practice (a polynomial analog of the Pocklington-Lehmer integer primality test) $\endgroup$ Jan 2, 2019 at 23:13
  • $\begingroup$ That's it. I spent awhile looking for a trick answer, but didn't see any. An alternative way of organizing the calculation would be to first build a list of remainders of $x^0,x^7\equiv-1+3x+3x^2+3x^3,x^{14}$ and $x^{21}$ modulo $f(x)$. Raising to seventh power modulo $f(x)$ is linear, so $$x^{49}\equiv-1+3x^7+3x^{14}+3x^{21}\pmod{f(x)}$$ et cetera. This may help with a higher degree polynomial. I'm afraid I don't have a good feel for the difference in complexity for I usually do this in characteristic two only :-) $\endgroup$ Jan 2, 2019 at 23:19
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You can also try to factor it as a product of two polynomials of degree $2$. Such a factorization must have the form

\begin{align}x^4 + 2x^2 - x + 1 & = (x^2+ax+b)(x^2-ax+b^{-1})\\ & =x^4+(b+b^{-1}-a^2)x^2+(ab^{-1}-ab)x+1\end{align} Hence the relations $$b+b^{-1}-a^2=2, ab^{-1}-ab=-1,$$ which can be easily seen to have no solution in $\mathbb{Z}_7$.

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