It is known that the set of primes $p$ which the quadratic polynomial $x^2+ax+b$ factors into linear factors $\pmod p$ (or over finite field of order $p$, $GF(p)$) is a set of modular congruences. For example, the set of primes $p$ which $x^2+2x-1$ completely factors over $GF(p)$ are of the form $8y+1$ or $8y+7$.
It is also known for any cubic polynomial $x^3+ax^2+bx+c$, the set of primes $p$ for which $x^3+ax^2+bx+c$ factors into linear factors $\pmod p$ either satisfy a set of modular congruences, or can be representable by a primitive integral binary quadratic forms of discriminant equal to the discriminant of the polynomial. For example, the set of primes $p$ for which $x^3-x-1$ completely factors over $GF(p)$ are of the form $p=y^2+23z^2$.
Let $P = x^4-x^3-2x^2-2x-1$. Then what is the set of primes $p$ such that $P$ completely factors into linear factors $\pmod p$, or over $GF(p)$? It is known that this set $S$ is a modular set (linear form), quadratic form, or cubic form. Thanks for help.
added. Caution: it is not a single quadratic form that represents the primes you want, it is two quadratic forms. (Monday) Indeed, we can take the two forms to be $$ x^2 + 95 y^2 \; , \; \; \; 5 x^2 + 19 y^2 \; , $$ as these represent the same odd numbers as $x^2 + xy + 24 y^2$ and $5 x^2 + 5 xy + 6 y^2;$ for either of these latter forms to be odd, we need $x(x+y)$ odd, therefore $x$ must be odd and $y$ must be even, leading to $y=2t$ and forms $(x+t)^2 + 95 t^2$ and $5(x+t)^2 + 19 t^2 \; .$
ORIGINAL:Quite surprised how this worked out. There is a 1973 article by Estes and Pall that proves that, for binary forms, the spinor kernel is the fourth powers in the form class group. At the end I have put a list up to 2000 of the relevant primes; 5 and 19 are not there, this is a simple program that just counts distinct roots mod p.
Your form discriminant is $-95,$ positive binary forms.
The forms in the principal genus that are not fourth powers are the pair of "opposites" $\langle 4,1,6 \rangle$ and $\langle 4,-1,6 \rangle.$ These represent the same primes, for which your polynomial splits as two irreducible quadratics. Added: if you prefer, you may use $\langle 9,4,11 \rangle,$ or $9x^2 + 4xy + 11 y^2,$ as it represents a subset of the same numbers, exactly the same odd numbers, and the same primes. Let's see: while it is necessary to allow $xy$ both positive and negative in searching for values of $9x^2 + 4xy + 11 y^2,$ we still get bounds on $|x|,|y|$ since $9x^2 + 4xy + 11 y^2 \geq \frac{95}{11} x^2$ and $9x^2 + 4xy + 11 y^2 \geq \frac{95}{9} y^2.$ The first few such primes are
The principal form is $\langle 1,1,24 \rangle.$ With these primes, your polynomial splits as four distinct linear factors. Added: if you prefer, you may use $\langle 1,0,95 \rangle,$ or $x^2 + 95 y^2,$ as it represents a subset of the same numbers, exactly the same odd numbers, and the same primes.
The other fourth power is $\langle 5,5,6 \rangle.$ This represents $5$ and $19,$ for which the polynomial has repeat roots. Other primes represented also split your polynomial into distinct linear factors. Added: if you prefer, you may use $\langle 5,0,19 \rangle,$ or $5x^2 + 19 y^2,$ as it represents a subset of the same numbers, exactly the same odd numbers, and the same primes.
Here is the form class group for discriminant $-95$
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These are the first few primes (NOT $5,19$ because they have repeated roots) for which the polynomial has four distinct roots.
Note that the version of the polynomial used on the field website is something like $ -x^4 \cdot f\left(\frac{-1}{x}\right)$
Added Monday lunchtime: perhaps a little more attractive to say the primes giving four linear factors are represented by the two forms $$ x^2 + 95 y^2 \; , \; \; \; 5 x^2 + 19 y^2 \; , $$ while the primes that give two irreducible quadratic are represented by $$ 9 x^2 \pm 4xy + 11 y^2 $$