Why is this polynomial irreducible Let $n \in \mathbb{N}_+$ and $a \in \mathbb{Q}$. Prove that the polynomial 
$$P(x) = x^{2^n}\left(x+a\right)^{2^n} + 1 \in \mathbb{Q}[x]$$
is irreducible.
I don't really have an idea what to do with it. For $n=1$ I could use brute-force approach, but this won't help. All my tries were in vain, and there were lots of them.
 A: Let's write $t=x+b$, $b=a/2$. Then $P=(t^2-b^2)^{2^n}+1 =\Phi(t^2-b^2)$, where $\Phi(x)=x^{2^n}+1$ is the (irreducible) cyclotomic polynomial of order $2^{n+1}$; its zeros are the primitive $2^{n+1}$st roots $\zeta$ of unity.
Thus the roots of $P$ are $\alpha=\pm\sqrt{b^2+\zeta}$, where $\zeta$ is as above. In particular, if we adjoin any such root, then $\mathbb Q (\alpha)\supseteq \mathbb Q(\zeta)$. The latter is an extension of degree $2^n$ of $\mathbb Q$. This shows that if we could factor $P=fg$, then $\deg f=\deg g= 2^n$ (because $f,g$ are multiples of the minimal polynomials of certain $\alpha$'s, and these have degree $\ge 2^n$, as we just saw). Also, we can now assume that $\mathbb Q(\alpha_1,\ldots, \alpha_{2^{n+1}})=\mathbb Q(\zeta)$; indeed, if $\mathbb Q(\alpha)\supsetneqq \mathbb Q(\zeta)$ for any $\alpha$, then we're done by the same argument.
In fact, we can be more specific. The Galois group of $\mathbb Q(\zeta)/\mathbb Q$ acts transitively on the $\zeta$, so $f(t)=\prod (t-\sqrt{b^2+\zeta})$ where $\zeta$ varies over all primitive roots, and $g$ has the zeros where we take the opposite sign when extracting the square roots. This implies that $g(t)=f(-t)$. So any non-trivial factorization of $P$ must take the form $P(t)=f(t)f(-t)$.
Now let's write $b=c/d$ and multiply through by $d^{2^{n+1}}$. Then our factorization becomes
$$
(d^2t^2-c^2)^{2^n}+d^{2^{n+1}}=F(t)F(-t), \quad F=d^{2^n} f .
$$
By Gauss's Lemma, $F\in\mathbb Z[x]$ now. But then comparing the constant terms gives that
$$
c^{2^{n+1}} + d^{2^{n+1}} = F_0^2.
$$
However, the equation $x^4+y^4=z^2$ has no integer solutions (an improvement on Fermat for $n=4$); see here.
This gives the claim except for the easy case $c=0$: then $b=0$ and $P$ becomes a cyclotomic polynomial.
