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).
Call $f$ your polynomial. Then $f$ is not irreducible if and only if there exists a root $x\in\mathbb{Q}$ of $f$. If a such root exists then, for $n\geq 1$ we have  $$0\leq(x(x-a)(x-b))^{2^n}=-1.$$

I leave the case $n=0$ to you.
Call $f$ your polynomial. Then $f$ is not irreducible if and only if there exists a root $x\in\mathbb{Q}$ of $f$. If a such root exists then, for $n\geq 1$ we have  $$0\leq(x(x-a)(x-b))^{2^n}=-1.$$

I leave the case $n=0$ to you.