Is $x^4+4$ an irreducible polynomial? 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. 
 A: $x^4+4 \cdot 1^4= x^4+ 2 \cdot 2 \cdot x^2+2^2 - (2x)^2$ 
Which is well known identity called Sophie Germain
A: As Berci showed, this polynomial is indeed reducible over the rationals.
One way to see it is to calculate its roots explicitly:
$$x^4+4=0 \leftrightarrow x^2 = \pm 2i \leftrightarrow x = \pm \sqrt{2} (\frac{\sqrt{2}}{2}(1+i)) \vee x = \pm  i\sqrt{2} (\frac{\sqrt{2}}{2}(1+i)) $$
Or:
$$x = \pm 1 \pm i$$
And since those roots are proper complex number in $\mathbb{Z}[i]$, you can pair $1+i$ with $\overline{1+i}=1-i$ and $-1+i$ with $\overline{-1+i} = -1-i$ and obtain the factorization $(x^2 - 2x + 2)(x^2 + 2x +2)$ (if $\alpha$ is a proper complex root of $p \in \mathbb{R}[x]$, then $\overline{\alpha}$ is another root, and $(x-\alpha)(x-\overline{\alpha}) = (x^2-2Re(\alpha) + |\alpha|^2)$ divides $p$.
A: $$X^4+4=X^4+4X^2+4-4X^2 =(X^2+2)^2-(2X)^2 \,.$$
A: $$\begin{eqnarray}x^4+4&=&(x^2+2i)\cdot (x^2-2i)\\ &=& (x-(1-i))\cdot (x+(1-i))\cdot (x-(1+i))\cdot(x+(1+i)) \\
&=& ((x-1)+i)\cdot ((x-1)-i)\cdot((x+1)-i)\cdot((x+1)+i) \\
&=& ((x-1)^2+1)\cdot((x+1)^2+1).\end{eqnarray}$$
Reducible.
A: One may use the same version of completing the square that proves that $x+\dfrac1x \ge 2$ when $x>0$:
$$
x+\frac1x = \left(x-2+\frac1x\right)+2 = \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2+2.
$$
Similarly
$$
x^4+4 = \left( x^4 +4x^2 + 4 \right) - 4x^2 = \left(x^2+2\right)^2 - (2x)^2
$$
then factor that as a difference of two squares.
A: 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 $\Bbb Q[x]$.
