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How to prove that polynomials of the form :

$P(x)= x^2+ax+a$ , where $a \in \mathbb{Z^{+}}$ \ $ \left \{ 4 \right \} $

are irreducible over ring $\mathbb{Z}$ of integers ?

Eisenstein's criterion and Cohn's criterion work fine for most of these polynomials but there are some exceptions that cannot be proved with these criterias such as irreducible polynomial:

$ x^2+36x+36$ or $ x^2+100x+100$ ,so I don't know how to prove that these exceptions are also irreducible polynomials .

I have checked statement above for many values of $a$ by small Maple program .

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up vote 5 down vote accepted

$P(x)$ is reducible in $\mathbb{R}[x]$: $$P(x)=\left(x-\frac{-a+\sqrt{a^2-4a}}{2}\right)\left(x-\frac{-a-\sqrt{a^2-4a}}{2}\right).$$

When $a\in \mathbb{Z}^+\setminus \{4\}$, $a^2-4a=(a-2)^2-4$ is not a square, so $\frac{-a\pm\sqrt{a^2-4a}}{2}$ is irrational. Hence, $P(x)$ is irreducible in $\mathbb{Z}[x]$.

(Proof that $(a-2)^2-4$ is not a square: suppose $(a-2)^2-4=k^2$ for some integer $k$. Then the two squares $(a-2)^2$ and $k^2$ differ by 4, which is possible only when $(a-2)^2=4$ and $k^2=0$. Therefore $a=0$ or $4$.)

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your proof is almost obvious...any hint how to prove that $x^4+x^3+x^2+ax+a$ is irreducible when $a$ is odd positive integer ? – pedja Nov 10 '11 at 12:57
@pedja I think in that case, the same method as posted by Bruno Joyal works (i.e. think in $\mathbb{F}_2[x]$). – pharmine Nov 10 '11 at 13:50
I will try that approach.. – pedja Nov 10 '11 at 13:54

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