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Show that $f(X)=3X^n+13X^{n-1}+11 $ is irreducible for $n \in \mathbb{N}_{\ge 2}$ over $\mathbb{Z}$. I tried to use Eisenstein's criterion but I couldn't find a $p$ which matches the criterion. I would be glad if you'd give me some advice.

Thank you.

maiT
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2 Answers2

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You can use the extended Eisenstein criterion to conclude that $f$ has an irreducible factor of degree at least $n-1$. But since $f$ has no integer roots, this already implies that $f$ is irreducible.

Below is said extended Eisenstein criterion, excerpted from here. enter image description here

Bill Dubuque
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Dietrich Burde
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If $f(X)=\left(\sum_{k=0}^p a_kX^k\right)\left(\sum_{k=0}^q b_kX^k\right)$ with $p$ and $q\geq 1$, you can assume WLOG $b_0=11$ and $a_0=1$.

Then $a_1b_0 + b_1a_0 = 0$, hence $11\mid b_1$. You also get $a_2b_0 + a_1b_1 + a_0b_2 = 0$, hence $11\mid b_2$. In a similar fashion, since the coefficient of $X^i$ in $f$ is $$\displaystyle \sum_{\ell=\max(i-p,0)}^{\min(i,q)}a_{i-\ell}b_\ell$$ we must have $11\mid b_i$ for all $i\leq \min(q,n-2)$.

  • In the case $q\leq n-2$, $11\mid b_q$ which cannot be, since $a_pb_q = 3$. Contradiction.

  • In the case $q> n-2$, since $p\geq 1$, we have $q=n-1$ and $p=1$, hence $b_{n-1}a_1=3$ and $b_{n-1}=13$, another contradiction.

Gabriel Romon
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  • This essentially reproves the extended Eisenstein criterion (cf. Dietrich's link) in this special case. Not a good idea pedagogically since the general proof is clearer and no more difficult. – Bill Dubuque Aug 05 '25 at 21:38