<s>[INST] On the reducibility of $x^4+ a$ over the rationals Let $a \in \mathbb{Z^{+}}$. Show that $x^4+a$ is reducible over $\mathbb{Q}$ if and only if $a=4b^4$ for some integer $b$.
My idea for one implication was to assume reducibility and write $x^4+a = (x^2+\alpha_1x + \beta_1)(x^2+\alpha_2x + \beta_2)$, leading to the system of equations: $\alpha_1 + \alpha_2 = 0$, $\beta_1+\beta_2+\alpha_1 \alpha_2 = 0$, $\alpha_2 \beta_1 + \alpha_1 \beta_2 = 0$, $\beta_1 \beta_2 = a$. But I wasn't able to get much farther than this. This is a practice problem for an exam, any help is appreciated.
 [/INST] Another solution is to first factor out into linear factors over $\mathbb{C}$, and multiply conjugate factors to obtain the real quadratic factors:
$$X^4 + a = \left(X^2 + ia^{1/2}\right)\left(X^2 - ia^{1/2}\right) = \color{red}{\left(X + \frac{1-i}{\sqrt{2}}a^{1/4}\right)}\color{blue}{\left(X - \frac{1-i}{\sqrt{2}}a^{1/4}\right)}\color{red}{\left(X + \frac{1+i}{\sqrt{2}}a^{1/4}\right)}\color{blue}{\left(X - \frac{1+i}{\sqrt{2}}a^{1/4}\right)}$$
$$ = \color{red}{\left(X^2 + \sqrt{2}a^{1/4}X + a^{1/2}\right)}\color{blue}{\left(X^2 - \sqrt{2}a^{1/4}X + a^{1/2}\right)}$$
where we used $\sqrt{i} = \pm \frac{1+i}{\sqrt{2}}$ and $\sqrt{-i} = \pm\frac{1-i}{\sqrt{2}}$.
(you can also check the factorization directly without using complex numbers).
From this expression it is clear that we need $\sqrt{2}a^{1/4} = 2r$ for some $r \in \mathbb{Q}$ which implies $a = 4r^4$.
</s>