Irreducibility of $X^4 + a^2$ with odd $a$ For my abstract algebra course I have to decide whether 
$$(*)\qquad  X^4 + a^2, \; a \in \mathbb{Z}\;\text{odd}$$ is irreducible over $\mathbb{Q}[X]$ and $\mathbb{Z}[X]$. 
Since the degree of the polynomial is 4 and thus even I should be able to split it into two polynomials of degree two $if$ the coefficients of those polynomials lie in $\mathbb{Z}[X]$ and/or $\mathbb{Q}[X]$ respectively. What I know is that, if I could show that it is irreducible over $\mathbb{Z}[X]$ then I would know by Gauss' Lemma it is so also over $\mathbb{Q}[X]$. On the other hand, I suspect that it might be reducible over $\mathbb{Q}[X]$ which, in turn, would imply it is reducible over $\mathbb{Z}[X]$. What I tried is to set up some general product of two second-degree polynomials 
$$(x^2 + bx + c)(x^2 + dx + e)$$ and get some conditions on which I could determine a set of parameters $\{b, c, d, e\}$ so that the polynomial in $(*)$ ensues. After coefficient comparison I get several conditions, i.e. 
$$b + d = 0 \\ 
  e + bd + c = 0 \\
  be + cd = 0 \\ 
  ce = a^2 = (2k + 1)^2
$$
for some $k \in \mathbb{Z}$.Here I did not know how to proceed. I saw that $b = -d$ and then tried to resolve the other equations but I did not make progress in that I found factors of (*). I obviously miss something here and appreciate a nudge in the right direction. 

Hint: the roots of $x^4 + a^2$ are
$$ \pm\sqrt{a}\frac{1 \pm i}{\sqrt{2}}. $$
Separate these into conjugate pairs to find the factorization of $x^4 + a^2$ over $\mathbf{R}[x]$.
If you want to save time, or perhaps check your work, the answer is
$$ x^4 + a^2 = (x^2 - \sqrt{2a} \cdot x + a)(x^2 + \sqrt{2a} \cdot x + a). $$
