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I'm trying to prove this question:

Show that $p(x)=x^3 + ax^2 + bx +1 \in \mathbb Z [x]$ is reducible over $\mathbb Z$ if and only if either $a=b$ or $a+b=-2$.

I did the converse in this way:

if we take $a=b$, we see easily that $p(-1)=0$, so $p(x)$ has a root over $\mathbb Z$, then $p(x)$ is reducible over $\mathbb Z$.

I'm having problems with the first implication, I need some hints.

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    Note that you're not done with the converse, because you haven't handled the case that $a+b = -2$.2012-10-20

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If $a+b = -2$, you can see that $p(1) = 0$. Now a polynomial of degree $3$ is reducible over $\mathbb Z$ if and only if it has a root. But you know by the rational root theorem that if $p(x)$ has a root over $\mathbb Q$, then that root, write it $q/r$, has to be such that $q \, | \, a_0$ and $r \, | \, a_3$, where $p(x) = a_3 x^3 + \dots + a_0$. It follows that the only two possibilities are $+1$ and $-1$.

The conditions $a=b$ or $a+b = -2$ correspond to the cases $+1$ and $-1$. I would've given an hint but stating the rational root theorem is pretty much giving the answer, so I thought I might as well do it.

Hope that helps,

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    It depends of the point of view, your question was perfect to me, because I'm studying this stuff right now in my Field Theory course. The fgp's answers was really interesting and elegant because even a student school could understand it. However your answer was more helpful to me, because of the arguments I said above.2012-10-20
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If $p$ is reducible then there are polynomials $q_1$, $q_2$ of lower degree such that $p=q_1q_2$. Now, since $p$ has degree 3, one of $q_1$,$q_2$ must have degree 2, and the other degree 1. Thus, if $p$ is reducible, there are constants such that $ p(x) = (a_2x^2 + a_1x + a_0)(b_1x + b_0) $

Since the coefficient of $x^3$ in $p$ is $1$, it must be that $a_2b_1 = 1$. And because the constant term of $p$ is $1$, you similarly get $a_0b_0 = 1$. Thus, if $p$ is reducible, it must have a zero at $1$ or $-1$. Which, in other words means $ \begin{eqnarray} a + b + 2 &=& 0 &\text{ or} \\ a - b &=& 0 \end{eqnarray} $

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    Because the sentence before that shows $b_1 \in \{-1,1\}$ and $b_0 \in \{-1,1\}$. That means $(b_1x + b_0)$ must have $1$ or $-1$ as a root.2012-10-20