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I want to find a 3rd degree polynomial in $\mathbb{Z}[X]$ that has no roots in $\mathbb{Z}$, but reducible in $\mathbb{Z}[X]$.

Tried to construct one as a product of two or three lower degree irreducible polynomials, but this won't work because one factor must be of the form $(x-r)$ where $r\notin\mathbb{Z}$.

Is there a general way to find such polynomials?

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    $\begingroup$ Try $(aX - b) P(X)$ where $P$ has degree two and is irreducible, and $a\wedge b = 1$ $\endgroup$ – Alexandre C. Oct 16 '16 at 12:27
  • $\begingroup$ @CaveJohnson great! thanks $\endgroup$ – Xena Oct 16 '16 at 12:28
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If the polynomial is reducible in $\mathbb{Z}[X]$ it is also reducible in $\mathbb{Q}[X]$; reducibility of a degree $3$ polynomial in $F[X]$, where $F$ is a field (including the case $F=\mathbb{Q}$), implies existence of a root in $F$.

So your polynomial must have a root in $\mathbb{Q}$ which is not in $\mathbb{Z}$. Can you find a degree $1$ polynomial in $\mathbb{Z}[X]$ that has no integer root? Then just cube it.

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