$x^4 + 2(1-a)x^2+ (1+a)^2$ is reducible over $\mathbb{Z}_p$.

Question

Let $a>1$ be an integer number and $p$ a prime number. Show that $p(x)=x^4 + 2(1-a)x^2+ 2(1+a)^2$ is reducible over $\mathbb{Z}_p$.

$\textbf{my attempt:}$

The easiest case is $p=2$, $p(x) = x^4 + 2(1+a)^2 = x^4 -(1+a)^2 = (x^2+(1+a))(x^2 + (1+a)).$ Then $p(x)$ is reducible over $\mathbb{Z}_p$.

There are another three cases, where $p \neq 2$:

1. $p|a$
2. $p$ doesn't divide $a$ and $a$ there exist $x \in \mathbb{Z}_p$ s.t $x^2=a$.
3. $p$ doesn't divide $a$ and $a$ there is no $x \in \mathbb{Z}_p$ s.t $x^2=a$. In this way we need to study if $-1$ has a root in $\mathbb{Z}_p$ or there is no $x \in \mathbb{Z}_p$ s.t $-1=x^2$.

I've solved the first case, if $p|a$ then $p(x)$ is $x^4 + 2x^2+1 = (x^2 +1)(x^2+1)$, because $1-a = 1$ and $(1+a)^2 = 1$ in $\mathbb{Z}_p$.

Since $p$ does not divide $a$ we have that $p$ does not divide $x$, then $a^\frac{p-1}{2}=1$, so we can write $$p(x) = x^4 + 2(a^\frac{p-1}{2}-a)x^2 + (a^\frac{p-1}{2}+a)^2$$.

But I can't see how to prove that $p(x)$ is reducible, can you help me? Also I need some help and tips in the third case.

Answer 1

The title is still inconsistent with your problem description. I will show $x^4+2(1-a)x+(1+a)^2$ is reducible modulo $p$.

There are only two possible factorizations (check why by yourself):

1. $(x^2+cx+d)(x^2-cx+d)$ where $c\ne 0\pmod{p}$
2. $(x^2+f)(x^2+g)$

Case 1: $p=4k+1$. In this case, $-1$ is a quadratic residue. Then let $c=2\sqrt{-1}$ and $d=-(1+a)$.

Case 2: $p=4k+3$. There are two sub-cases.

Case 2.1: $a$ is a quadratic residue. Let $d=1+a$ and $c=2\sqrt{a}$.

Case 2.2: $a$ is a nonresidue. Then it can be factorized as $(x^2+f)(x^2+g)$. To see this, we need to check $t^2+2(1-a)t+(1+a)^2=0$ has two roots (you can think of $t=x^2$). Let $\Delta=(2(1-a))^2-4(1+a)^2=-16a$. We need to check $\sqrt{\Delta}$ exists, i.e., $\Delta$ is a quadratic residue. Recall that $-1$ is also a nonresidue in this case. Then $-16a$ is a residue. This is because modulo a prime, the product of two nonresidues is a residue.