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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
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5 Answers

up vote 27 down vote accepted

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

Reducible.

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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
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@Sigur: no general one, because that would probably give you an efficient algorithm for prime factorization. –  akkkk Dec 8 '12 at 14:05
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I edited your answer so that it wasn't all on one line. –  Fredrik Meyer Dec 8 '12 at 14:16
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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
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$x^4+4 \cdot 1^4= x^4+ 2 \cdot 2 \cdot x^2+2^2 - (2x)^2$

Which is well known identity called Sophie Germain

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

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