Find the set of primes $p$ (either a modular congruence or quadratic form) such that $x^3-4x-1$ factors into three linear factors over the field $GF(p)$. I noted that these primes $p$ include $37, 53, 173, 193, 229, 241$. For example,
$x^3-4x-1 = (x + 8)(x + 13)(x + 16)$ in the finite field of order $37$, $GF(37)$ hence $37$ is one of these primes in the set.
I would expect some statement such as $x^3-4x-1$ factors into only linear factors over the finite field $GF(p)$ if and only if $p$ is of the form $x^2+ay^2$ for some integer $a$, or if $p$ holds a specific modular congruence.
In A Course in Computational Algebraic Number Theory by Henri Cohen, see Appendix B.4, Table of Class Numbers and Units of Totally Real Number Fields, pages 521-523. Apparently I made a jpeg of this years ago.
The material underlying my conclusion is the main Theorem in Spearman Williams (1992)
If you wish, you may use $$ p = x^2 - 229 y^2 $$
First Version: $$ p = x^2 + 15 xy - y^2 $$
Represented (positive) primes up to 10000
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Next are two pages from the 1976 article by Daniel Shanks, which states the theorem about cubic fields without proof. It is the paragraph on page 29 that begins "The general rule is simply this:"