Let $n=2^s$ with $s \geq 1$. Is the polynomial $$(1+x)^n+(1-x)^n \in \mathbb{Q}[x]$$ irreducible? I have checked this for some values of $s$ and it seems to be true. Notice that we can also write it as $2 \sum_{k \geq 0} \binom{n}{2k} x^{2k}$. I have already found an irreducible factorization with substituted cyclotomic polynomials if $n$ is odd. The even case leads to the case of $n=2^s$.
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Your polynomial $P_n(x)$ is irreducible if and only if $Q_n(z) = z^n + 1$ is irreducible (the fractional linear transformation $z = (1+x)/(1-x)$ maps from one to the other). It is well-known that $Q_{2^n}$ is the cyclotomic polynomial $\Phi_{2^{n+1}}$, and thus is irreducible.
EDIT: In general, consider a fractional linear transformation $$z = \phi(x) = \dfrac{ax+b}{cx+d}$$ where $a,b,c,d$ are rational, $ad-bc \ne 0$, $c \ne 0$. For a polynomial $P$ of degree $m$ over $\mathbb Q$, we have
$$P(z) = (cx+d)^{-m} R(x)$$ where $R$ is again a polynomial of degree $\le m$. Since $P(a/c) = \lim_{x \to \infty} P(\phi(x))$, $R(x)$ has degree $m$ as long as $P(a/c) \ne 0$. In the other direction, $\phi^{-1}(z) = \dfrac{dz-b}{-cz+a}$, and $cx+d = (ad-bc)/(a-cz)$ where $z = \phi(x)$, so that $$P(z) = (ad-bc)^{-m} (a-cz)^m R(\phi^{-1}(z))$$ so any polynomial $R$ of degree $m$ such that $R(-d/c) \ne 0$ arises in this way from such a polynomial $P$. Now $P$ factors as $P_1 P_2$ with degrees $m_1, m_2$ iff $R$ factors as $R_1 R_2$ with degrees $m_1, m_2$, where $P_i(z) = (cx+d)^{-m_i} R_i(x)$.
Robert Israel
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