Proposition: The polynomial (x^4+1) is reducible in the ring (\mathbb Z/p \mathbb Z)[x] for every prime p.


If p=2 then x^4+1 = (x+1)^4 is clearly reducible. Otherwise, let p be an odd prime. Then, p^2\equiv 1 (\mod 8), so

(x^4+1) | (x^8-1) | (x^{{p^2}-1}-1) | (x^{p^2}-x)

Hence the roots of the given polynomial lie in the field \mathbb F_{p^2}. Had the quartic been irreducible, the roots would have been found in an extension of degree 4 over \mathbb F_p.

Proposition: The polynomial (x^2-2)(x^2-3)(x^2-6) has a root modulo every prime p, that is, in (\mathbb Z/p \mathbb Z)[x].


If \displaystyle\left(\frac{2}{p}\right) = \left(\frac{3}{p}\right) = -1, then by multiplicativity of the Legendre symbol, \displaystyle\left(\frac{6}{p}\right) = 1.