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Is the cubic polynomial $x^3-ax-(2a+1)$ irreducible over $\mathbb Z$ for all positive integers $a$?

one way to prove irreducibility is to use eisenstein's criterion. we want to find prime $p$ st:
$p\mid a$
$p\mid 2a+1$
$p^2\nmid 2a+1$
However, the first 2 conditions imply $p\mid 1$, a contradiction. so this approach fails.

another approach is to use the rational root thm.
let $x^3-ax-(2a+1)=(x+b)(x^2+cx+d)=x^3+(b+c)x^2+(bc+d)x+bd$
where b,c,d integers
$b+c=0$
$bc+d=-a$
$bd=-(2a+1)$

there was a link to a similar problem, now deleted.

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  • $\begingroup$ Your current question as well as the linked question do not have any context. Please try to improve your question by adding relevant context. $\endgroup$
    – Paramanand Singh ♦
    Jan 12, 2022 at 2:56
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    $\begingroup$ No!! Context has to be part of the question body. I understand that you have acted here in good faith but note that while self answering the bar for standards is higher. For the current situation it is best to close the question unless improved. $\endgroup$
    – Paramanand Singh ♦
    Jan 12, 2022 at 3:08

1 Answer 1

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If a cubic is reducible, then it must have a linear factor $x+b$, so $-b$ is an integer root.

$$-b^3+ab-(2a+1)=0$$ $$a(b-2)=b^3+1$$ $$a=\frac{b^3+1}{b-2}$$ $$(b-2)\mid 9$$ \begin{align}b-2&=-9,-3,-1,1,3,9\\b&=-7,-1,1,3,5,11\\b&\neq -1,1\end{align} so $(a,b,c)=(38,-7,11),(28,3,-19),(42,5,-17),(148,11,-27)$.

\begin{align}(x-7)(x^2+7x+11)&=x^3-38x-77\\ (x+3)(x^2-3x-19)&=x^3-28x-57\\ (x+5)(x^2-5x-17)&=x^3-42x-85\\ (x+11)(x^2-11x-27)&=x^3-148x-297\end{align}

so there are 4 $a$'s such that the cubic is reducible. Simply stating (but not deriving) one of the 4 factorizations is enough to disprove the question.

thanks to @dxiv for the suggestion.

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    $\begingroup$ Nice answer (+1), could use a bit more explanation at a few points. $\endgroup$
    – rogerl
    Dec 18, 2021 at 19:26
  • $\begingroup$ Can you elaborate on why $b-2$ must divide $9$? $\endgroup$
    – 2'5 9'2
    Dec 18, 2021 at 19:55
  • $\begingroup$ I see, because $a=\frac{b^3+1}{b-2}=b^2+2b-4+\frac{9}{b-2}$. $\endgroup$
    – 2'5 9'2
    Dec 18, 2021 at 20:01
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    $\begingroup$ You can shorten the proof somewhat by noting that any rational root must be an integer since the polynomial is monic. Let $n$ be such a root, then $a=\frac{n^3-1}{n+2}$ and a similar argument follows. $\endgroup$
    – dxiv
    Dec 18, 2021 at 22:04
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    $\begingroup$ @cineel Regroup and collect: $\,n^3-an-(2a+1)=0 \iff n^3-1=(n+2)a\,$. $\endgroup$
    – dxiv
    Dec 19, 2021 at 5:34

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