TITLE: Is $x^4+4$ an irreducible polynomial?
QUESTION [10 upvotes]: We know that $p(x)=x^4-4=(x^2-2)(x^2+2)$ is reducible over $\mathbb{Q}$ even not having roots there. 
What about $q(x)=x^4+4\in \mathbb{Q}[x]$? Again, no roots.

REPLY [2 votes]: This is Sophie Germain's result(don't you think her name is worth mentioning here). If this was Gauss's result, someone would have mention the name Gauss. But this is not.

Sophie Germain Identity
  \begin{align}
x^4+4y^4= (x^{2}+2y^{2}-2xy)(x^{2}+2y^{2}+2xy)\tag{1}
\end{align}

To answer the question,
Put $y=1$ in $(1)$. Thus
$$x^4+4= (x^{2}+2-2x)(x^{2}+2+2x)$$
Hence $q(x)$ reducible over $\Bbb Q[x]$.