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Show that the polynomial $x^3+nx+2$ is irreductible in $\mathbb{Z}[x]$ for all integers $n \in \mathbb{Z}-\{1,-3,-5\}$.

If $n$ is even, then $2 | n$, $2|2$ and $2^2 \not| 2$, then by Eisenstein's criterion, $x^3+nx+2$ is irreductible.

If $n$ is odd and $n \in \mathbb{Z}-\{1,-3,-5\}$, then find $a \in \mathbb{Z}$ and a prime number $p$ respecting the conditions of Eisenstein's criterion for $(x+a)^3+n(x+a)+2=x^3+3ax^2+(3a^2+n)x+(a^3+na+2)$.

Which $a$ and $p$ could I take here?

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Let $p(x)=x^3+nx+2$. $p$ is reducible in $\Bbb Z[x]$ iff it has a root in $\Bbb Z$

By the rational root theorem, the possible roots are $\pm 1, \pm 2$.

$$p(1)=n+3=0\iff n=-3\\ p(-1)=-n+1 =0\iff n=1\\ p(2) = 10+2n=0 \iff n=-5\\ p(-2) = -6-2n=0 \iff n=-3$$

Can you go on?

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  • $\begingroup$ However, if $p(-1)=0$, then $n=1$. $\endgroup$ – Taj Mohamed Bandalandabad Mar 10 '16 at 0:07
  • $\begingroup$ @TajMohamedBandalandabad True, sorry! $\endgroup$ – YoTengoUnLCD Mar 10 '16 at 0:11

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