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So I have seen the similar question and answers on here for $x^4 +1$, but I am having trouble extending anything there to this polynomial... I understand it is fairly trivial with Galois theory, but my class has just barely covered Field Extensions, so suffice it to say we have no Galois theory to play with.

I managed to prove it for the primes such that $p \equiv 1, 7 \pmod 8$, by noting that $2$ is a square modulo those primes and thus $x^4 - 2x^2 +1 = (x^2 -1 + 2qx)(x^2 - 1 - 2qx)$ for those $\mathbb{Z}_p$... however, trying to get a similar result for $3 \pmod 8$ and $5 \pmod 8$ has been stumping me for a long time, I am having a hard time making $q^2 = -1$ and $q^2 = -2$ give me something factorable...

I guess the worst part of all of this is that I don't think this solution is even particularly enlightening, in terms of abstract algebra. It's really just some number theory trickery. I don't think my course has prepared me theoretically for this problem, does anyone have an elementary approach to it?

user44400
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abhalphiest
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2 Answers2

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We have $$x^4-10x^2+1 =(x^2-1)^2-8x^2 = (x^2-1-2ax)(x^2-1+2ax)$$ if $2=a^2$, or $$x^4-10x^2+1 =(x^2+1)^2-12x^2 =(x^2+1-2bx)(x^2+1+2bx)$$ if $3=b^2$. Thus the only cases that remain are those where neither $2$ nor $3$ is a square - but then $6$ is a square and with $6=c^2$, we find $$x^4-10x^2+1 =(x^2-5)^2-24=(x^2-5-2c)(x^2-5+2c).$$

(Note that we do not even need quadratic reciprocity or similar "advanced" stuff - all we need is that the product of non-squares is square, which follows from the fact that $\Bbb F_p^\times$ is cyclic).

Hagen von Eitzen
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Another simple explanation comes from the fact that the zeros of $m(x)=x^4-10x^2+1$ are $$ x=\pm\sqrt2\pm\sqrt3 $$ with all four sign combinations. So if $p$ is a prime, then you get the splitting field of $m(x)$ over $K=\Bbb{F}_p$ by adjoining $\sqrt2$ and $\sqrt3$. Because up to isomorphism the field $K$ has only a single quadratic extension, namely $L=\Bbb{F}_{p^2}$, we immediately see that $m(x)$ splits into linear factors over $L$. This is because $\sqrt2$ and $\sqrt3$ are both elements of $L$. Consequently $m(x)$ splits into quadratics at worst over $K$.

From this way of looking at it it is obvious how to generalize this. Any biquadratic polynomial with zeros of the form $\pm\sqrt{d_1}\pm\sqrt{d_2}$ for some integers $d_1,d_2$ will split into quadratic (or linear) factors modulo $p$ for all primes $p$. From basic field theory we see that it will be irreducible over $\Bbb{Q}$ whenever the field $F=\Bbb{Q}(\sqrt{d_1},\sqrt{d_2})$ is a degree four extension over $\Bbb{Q}$.

See Qiaochu's answer here for more Galois theory.

Jyrki Lahtonen
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    Of course, the fact that by adjoining a single square root to $\Bbb{F}_p$ you get all of them also follows from the observation that modulo $p$ the product of two quadratic non-residues is a quadratic residue. – Jyrki Lahtonen Aug 01 '16 at 06:47
  • Why can you say √3-√2 is also root of m（x）？（of course √２＋√３　is a root of m（x）by definiton, but what about √3-√２？） – Pont Jan 28 '21 at 17:31
  • Why m（x）spilts over quadratics at worst over K? I see √２＋√３　is a root of m（x）,but we cannot say m（x） spilts useless some other 1-degree factor exists. – Pont Jan 29 '21 at 06:00
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    @bellow The roots of $x^4-10x^2+1$ are $\pm\sqrt2\pm\sqrt3$. Both $\sqrt2\in L$ and $\sqrt3\in L$. Therefore all the roots are in $L$, and the quartic splits over $L$. On the other hand, all the elements of $L$ have minimal polynomials that are at worst quadratic over the prime field $K$. So the quartic factors of the prime field. It depends on the choice of $p$ exactly how it factors, but the point of my argument was to avoid splitting this in to cases according to which of $2,3,6$ are quadratic residues modulo $p$. – Jyrki Lahtonen Jan 29 '21 at 16:01