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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.

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Notice that $x^4 + 4 = (x^2+2)^2 - (2x)^2$. – Erick Wong Dec 8 '12 at 19:13
up vote 29 down vote accepted

$$\begin{eqnarray}x^4+4&=&(x^2+2i)\cdot (x^2-2i)\\ &=& (x-(1-i))\cdot (x+(1-i))\cdot (x-(1+i))\cdot(x+(1+i)) \\ &=& ((x-1)+i)\cdot ((x-1)-i)\cdot((x+1)-i)\cdot((x+1)+i) \\ &=& ((x-1)^2+1)\cdot((x+1)^2+1).\end{eqnarray}$$


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Great and thanks. Just one more question: is there some irreducibility criterion to answer this without trying to factor it? – Sigur Dec 8 '12 at 13:57
@Sigur: sometimes you can use Eisenstein's criterion, but it is generally a hard problem. – akkkk Dec 8 '12 at 14:03
@akkkk, it does not apply here. Any else? – Sigur Dec 8 '12 at 14:04
@Sigur: no general one, because that would probably give you an efficient algorithm for prime factorization. – akkkk Dec 8 '12 at 14:05
I edited your answer so that it wasn't all on one line. – Fredrik Meyer Dec 8 '12 at 14:16

$x^4+4 \cdot 1^4= x^4+ 2 \cdot 2 \cdot x^2+2^2 - (2x)^2$

Which is well known identity called Sophie Germain

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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 $p$.

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Amazing how a simple question can produce many interesting facts. – Sigur Dec 8 '12 at 14:20

$$X^4+4=X^4+4X^2+4-4X^2 =(X+2)^2-(2X)^2 \,.$$

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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.

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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 $\Bbb Q[x]$.

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See below. – Sigur Oct 30 '15 at 15:08
Oh, I don't know how I missed that. Thanks! – albo Nov 1 '15 at 12:45

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