Prove that $x^4 + ax^2 + b^2$ is reducible over $\mathbb{Z}_p$ with $p$ prime. My work
$$x^4 + ax^2 + b^2 = (x\pm b)^2 + (a \mp 2b)x^2$$
Then if either $(a + 2b) = -r^2$ or $(a - 2b) = -q^2$ with $q,r \in \mathbb{Z}_p$ the result follows.
But I'm not really sure how to prove this last part.
I'd really appreciate any help you could give me.
 A: The key point is to get inspiration from factoring that polynomial over the complex numbers into two quadratic factors (with leading coefficient 1).  Use the quadratic formula to work out $x^2$ and then $x$, so you find all four complex roots $r_1, r_2, r_3, r_4$.  Therefore the quartic in your question factors into two quadratics in three ways:
$$
(x-r_1)(x-r_2) \cdot (x-r_3)(x-r_4), \ \ (x-r_1)(x-r_3) \cdot (x-r_2)(x-r_4), \ \ (x-r_1)(x-r_4) \cdot (x-r_2)(x-r_3).
$$
If you work out what the six possible quadratic factors (such as $(x-r_1)(x-r_2)$ and $(x-r_3)(x-r_4)$) really are, you'll see they involve either $\sqrt{a}$ or $\sqrt{b}$ or $\sqrt{ab}$.  All of this is purely algebraic, and that means that as long as $a$ or $b$ or $ab$ is a square in some field, then the factorization involving that square root will make sense using coefficients in that field.  So we finally reach the key point: for all $a$ and $b$ in the integers modulo a prime $p$, either $a$ or $b$ or $ab$ is a square.
