# Is $\bigl(X(X-a)(X-b)\bigr)^{2^n} +1$ an irreducible polynomial over $\mathbb{Q}[X]$?

Let $a, b \in \mathbb{Q}$, with $a\neq b$ and $ab\neq 0$, and $n$ a positive integer.

Is the polynomial $\bigl(X(X-a)(X-b)\bigr)^{2^n} +1$ irreducible over $\mathbb{Q}[X]$?

I know that $\bigl(X(X-a)\bigr)^{2^n} +1$ is irreducible over $\mathbb{Q}[X]$, but I have a hard time generalizing my proof with three factors.

PS: This is not homework (and may even be open).

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Why is this tagged "cyclotomic polynomial"? –  Zev Chonoles Jul 2 '13 at 2:44
Cool question, got me thinking. Would you mind explaining how you know that ${(X(X - a))}^{2^n} + 1$ is irreducible over Q? –  Robert Lewis Jul 2 '13 at 2:51
Robert, not sure I can detail enough in the comment box, but if $a=0$ it's trivial (cyclotomic), suppose $a\neq 0$, let $z \in \mathbb{C}$ a root, and $\mathbb{Q}[z]$ is a field extension of $\mathbb{Q}[z(z-a)]$. Looking at the minimal polynomial of $z$ over $\mathbb{Q}$ we show that $\mathbb{Q}[z(z-a)] \neq \mathbb{Q}[z]$ and we conclude with the fact that $\mathrm{dim}_{\mathbb{Q}}\mathbb{Q}[z(z-a)]=2^n$. –  user84673 Jul 2 '13 at 3:09
This is far from an open question since it is a special case of an old conjecture of Schur proved long time ago by Seres. If I understand well you want to use Capelli's Lemma in order to prove the irreducibility. You said (but didn't show us!) that you were succesful for the product of two distinct polynomials $X$ and $X-a$. I'm eager to see a proof on this line at least for some particular cases like these. (I don't know a proof of Schur conjecture using Capelli's Lemma, if there is one.) –  user26857 Jul 2 '13 at 17:59
Schur's conjecture was the following: if $f(X)=X^{2^n}+1$, $n\ge 1$, and $g(X)=(X-a_1)\cdots(X-a_m)$ with $a_i\in\mathbb Z$ distinct, then $f(g(X))$ is irreducible over $\mathbb Q$. –  user26857 Jul 2 '13 at 19:05