x^3 + 3x + 1 splits mod p if and only if p can be represented by one of the forms a^2 + ab + 34b^2, 5a^2 + 5ab + 8b^2.
P(x) = x^4 - x^3 - 7*x^2 + 2*x + 9
splits modulo a prime p if and only if p can be written in the form
F(a,b,c) = a^3 + b^3 + c^3 + 11*(a^2*b + b^2*c + c^2*a) - 14*(b^2*a + c^2*b + a^2*c) - 157*a*b*c
for integers a, b, c.
If p is not a cube mod 163, then P(x) factors as a linear factor times an irreducible cubic modulo p, while if p is a cube mod 163 but not representable in the form F(a,b,c) then P(x) factors as a product of two quadratic polynomials modulo p. (Modulo 163, we have F(a,b,c) = (a + 58*b - 59*c)^3.)
Mod 2, P(x) = (x + 1)*(x^3 + x + 1), and 2 is not a cube modulo 163.
Mod 5, P(x) = (x^2 + x + 1)*(x^2 - 2*x - 1), and 5 = 32^3 (mod 163), but 5 isn't represented by F(a,b,c).
Mod 241, P(x) = (x + 41)*(x + 45)*(x - 68)*(x - 19), and 241 = F(1, 2, -1), so 241 = (1 + 2*58 - (-1)*59)^3 = 13^3 (mod 163).
Any monic cubic polynomial which doesn't have any roots in Q2 or Q3 has a root in the noncommutative ring
R = Z〈a,x〉/(a^3 = a+1, x^3 = 3x-1, xa = a(x^2-2)-2+x).
The equation s^3 = 2 has the solution s = -1-a-xa in R. This s also satisfies the commutation relation sx = (-x^2-x+2)s.
R is a maximal order inside the central simple algebra over Q with invariant 1/3 at 3 and 2/3 at 2. R has (left) class number 1 and discriminant -6^6.
If we set b = sas-1, c = s-1as, then the norm one part of the unit group has the presentation 〈a, b, c | ab3 = ca2c, bc3 = ab2a, ca3 = bc2b, a2c2 = b2ab, b2a2 = c2bc, c2b2 = a2ca, ab2a3 = cbab3, bc2b3 = acbc3, ca2c3 = baca3, a5b3a3b2 = 1〉. The outer automorphism group is generated by conjugation by s, which sends a→b, b→c, c→a, and conjugation by 1+s, which sends a→bc2, b→ca2, c→ab2. The cohomology groups of the norm one unit group are: H0 = Z, H1 = 0, H2 = Z/91, H3 = Z/2, H4 = Z/91, H5 = Z. The cup product H2 ⊗ H2 → H4 is an isomorphism.