Irreducibility of $x^4-32x^3+240x^2-320x-368=0$ over $\mathbb{Z}$

Question

Prove that $x^4-32x^3+240x^2-320x-368=0$ is irreducible over $\mathbb{Z}$

My working :I've tried using Perron's irreducibility criterion but not applicable here, tried Cohn's irreducibility criterion and also tried Eisenstein's criterion but could not found any prime $p$ \begin{align*} 32&=2^5\\240&=2^4\times 3\times 5\\320&=2^6\times 5\\368&=2^4\times 23\end{align*}

Meanwhile, reducible if and only if a translate is reducible, take $x = t + 8$ and get $t^4 - 144 t^2 -576 t + 144$ The possible factorings are all $ (t^2 + bt + c)(t^2 - bt + d)$ where $cd = 144$ — Will Jagy, May 22 '22 at 15:43

score 4 · Accepted Answer · answered May 22 '22 at 15:48

Try some substitutions. First: since you have coefficients involving high powers of $2$, put in $x=2y$. After dividing out a common factor of $16$ you get

$y^4-16y^3+60y^2-40y-23=0.$

This is almost $2$-Eisenstein, except the final coefficient is odd.

To remove that defect, try an odd increment, $y=z+\text{an odd number}$. I find that $y=z+1$ gives an equation for $z$ that is fully $2$-Eisenstein, and irreducibility is proved.

Dietrich Burde · Answer 2 · 2022-05-22T16:39:50.103

The direct way seems to be the best here, since the polynomial is also reducible modulo $p$ for every prime $p$. Since $f$ has no rational root, we may assume that $$ f=(x^2+ax+b)(x^2+cx+d) $$ with integral coefficients. By comparison we obtain equations over the integers with no solution: \begin{align*} 0 & = a+c+32,\\ 0 & = ac + b + d - 240,\\ 0 & = ad + bc + 320,\\ 0 & = bd+368. \end{align*}

It follows that $f$ is irreducible in $\Bbb Z[x]$ and hence in $\Bbb Q[x]$.

Irreducibility of $x^4-32x^3+240x^2-320x-368=0$ over $\mathbb{Z}$

2 Answers2