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Let $h(x)=x^4+12x^3+14x^2-12x+1$, and let $p>5$ be a prime.

I want to show $h(x)$ factors into 2 quadratics mod $p$ if $p \equiv 9,11$ mod 20, while $h(x)$ factors mod $p$ into 4 linear factors if $p \equiv 1,19$ mod 20. I can show $h(x)$ is irreducible if $p \equiv 3,7$ mod 10.

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This has a homework scent to it... – Igor Rivin Mar 14 '12 at 18:23
Can you tell us the Galois group of $h$ over $\mathbf{Q}$? It must be abelian if the splitting is determined by congruence conditions. – Timo Keller Mar 14 '12 at 18:26
MAGMA calculated the Klein four group $\mathbf{Z}/2 \times \mathbf{Z}^2$. Now use class field theory and Chebotarev. – Timo Keller Mar 14 '12 at 18:47
It seems the congruences you want are $p \equiv \pm 11 \mod 30$, and $p \equiv \pm 1 \mod 30$. The other cases $(7,13,17,23)$ are taken care of. – Zack Wolske Mar 14 '12 at 23:10
This was also posted on math stackexchange. – Franz Lemmermeyer Mar 15 '12 at 5:13
up vote 5 down vote accepted

It's been noted already that in fact $h \bmod p$ has four linear factors iff $p \equiv \pm 1 \bmod 30$, and is a product of two quadratics iff $p \equiv \pm 11 \bmod 30$. This can be checked by identifying the splitting field of $h$ with the real subfield of the $15$-th cyclotomic field, generated by $c := e^{2\pi i/15} + e^{-2\pi i/15} = 2 \cos (2\pi/15)$ which is a root of $c^4 - c^3 - 4c^2 + 4c + 1 = 0$; indeed $1 + 2(c-c^2)$ is a root of $h$. The desired result soon follows from the fact that Frobenius takes $e^{2\pi i/15}$ to $e^{2p\pi i/15}$. This is consistent with a cyclic Galois group (not the Klein 4-group as some have claimed), since ${\rm Gal}({\bf Q}(c)/{\bf Q})$ is the cyclic group $({\bf Z} / 15 {\bf Z})^* / \lbrace \pm1 \rbrace$.

If for some reason you do need a quartic of this form $x^4 + ax^3 + bx^2 - ax + 1$ (i.e. with a symmetry $x \leftrightarrow -1/x$) that splits completely mod $p$ iff $p \equiv \pm 1 \bmod 20$, the first few possibilities are $(a,b) = (\pm 2,-6)$, $\pm(22, -6)$, and $(\pm 18,74)$ if I computed correctly in gp.

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NOTE: I am doing $x^4 + 12 x^3 + 14 x^2 - 12 x + 1.$

Something is very wrong, perhaps just the typing of the polynomial. With gp-pari, I do get irreducible $\pmod p$ for primes $p \equiv \pm 3 \pmod {10},$ after $$(x+1)^4 \pmod 2, \; \; (x^2 + 1)^2 \pmod 3, \; \; (x+3)^4 \pmod 5. $$

Pari says the discriminant is $2^{12} \cdot 3^2 \cdot 5^3.$

It gives four linear factors for $$ p \in \{ 29, 31, 59, 61, 89, 149, 151, 179, 181, 211, 239, 241, \ldots \} $$

It gives two quadratic factors for $$ p \in \{ 11,19,41,71,79,101,109,131,139,191,199,229,251, \ldots, 409,\ldots \} $$

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The Galois group of the splitting field of $h$ over $\mathbf{Q}$ is $\mathbf{Z}/2 \times \mathbf{Z}/2$, the Klein four group. Since it is abelian, the splitting of primes is determined by congruence conditions, and the density of the splitting types is given by Chebotarev. You can realise the splitting field as a subfield of $\mathbf{Q}(\zeta_{\mathrm{|disc|}})$ and use what you know about splitting of primes in cyclotomic extensions.

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The theory mentioned no doubt explains these observations about The quartic $x^4 + 12 x^3 + 14 x^2 - 12 x + 1$ for primes $7 \le p \le 997$:

It is irreducible for $p \equiv \pm 2 \mod{5}$ but close to equally split between 2 and 4 factors for primes $p \equiv\pm 1\mod{5}.$ Even $\mod 80$ there is about an even split.

However, it is irreducible for $p \equiv\pm 2,\pm 7 \mod{15}$, has two quadratic factors for $p \equiv \pm 4 \mod{15}$ and four linear factors for $p \equiv\pm 1 \mod{15}.$

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For $p> 5$, suppose $p \equiv \pm 1$ or $\pm 11 \mod 30$. Then $p \equiv \pm 1 \mod 5$, so by reciprocity $5$ is a quadratic residue. Let $r^2 \equiv 5.$ Then for $a = 6+2r$ and $b=6-2r$, $(x^2 + ax -1)(x^2 + bx -1) \equiv x^4+12x^3+14x^2-12x+1 \mod p$.

To reduce to linear terms, we require $(x + (3 \pm r))^2 \equiv 1 + (3 \pm r)^2$, so we need $15 \pm 6r$ to be quadratic residues. Note that this is $3r(r \pm 2)$, and that $(r+2)(r-2) \equiv 1$, so $15 + 6r$ is a residue iff $15 - 6r$ is (i.e. the quadratics given either both split or are both irreducible). Perhaps someone else has a good idea to close this direction; I'd need at least another coffee.

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