Is $y^3 + ny + 1$ is reducible over $\mathbb{Z}$ for infinitely many $n $? Yes/NO

Question

Given $n \in \mathbb{Z}$

Is $y^3 + ny + 1$ is reducible over $\mathbb{Z}$ for infinitely many $n $? Yes/NO

My attempt : I thinks yes Every odd degree polynomial has atleast one roots

Is its true ?

Answer 1

Yes it is irreducible for all $n$ satisfying $|n| > 2$.

If it were not, then $y^3+ny+1$ would have an integral root $y_0$. But this is impossible, as $y^3_0$ is a multiple of $y$ and $ny_0+1$ is not for every integer $y_0 \not = \pm 1$. But for $y_0 = \pm 1$ note that $|n|$ must be no larger than 2.

Answer 2

If $y$ is an integer root, so $y^3+1$ is divisible by $y$, which gives not so many cases.

Answer 3

By the Rational Root Theorem, the only possible rational roots are $\pm 1$, so that $(\pm 1)^3 + n (\pm 1) + 1 = 0$. Rearranging gives $\mp(n + 1) = 1$, leaving at most two values of $n$ for which the polynomial is reducible.