1.
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.

2.
The polynomial

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.)

Examples:
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).

3.
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).

Example:
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.