Reducibility of $x^{2n} + x^{2n-2} + \cdots + x^{2} + 1$ Just for fun I am experimenting with irreducibility of certain polynomials over the integers. Since $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$, I thought perhaps $x^6+x^4+x^2+1$ is also reducible. Indeed:
$$x^6+x^4+x^2+1=(x^2+1)(x^4+1)$$
Let $f_n(x)=x^{2n}+x^{2n-2}+\cdots + x^2+1$. Using Macaulay2 (powerful software package) I checked that:
$$f_4(x) = (x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$$
 We get that
$$ f_5(x)=(x^2+1)(x^2-x+1)(x^2+x+1)(x^4-x^2+1)$$
The polynomial is reducible for $n=6, 7, 8, 9$ as far as I checked. I suspect that $f_n(x)$ is reducible over integers for all $n\ge 2$. Is this true?
Thanks!

Let $g_n(x) = x^n + x^{n-1} + \ldots + x + 1$. Then $g_n(x) = \frac{x^{n+1}-1}{x-1}$. Your polynomial $f_n(x) = g_n(x^2)$. The factorization of $x^n-1$ over $\mathbb{Q}$ is known:
$$
x^n-1 = \prod_{d\mid n}{\Phi_d(x)},
$$
where $\Phi_d(x)$ is the $d$th cyclotomic polynomial. Thus, the factorization of $g_n(x)$ will lead to a factorization of $f_n(x) = g_n(x^2)$ in $\mathbb{Q}[x^2]$. It is then possible that this factorization may be refined further. So, the only chance of an irreducible $f_n(x)$ would be when $n = p-1$ for some prime: in this case, it is known that $g_{p-1}(x) = x^{p-1} + \ldots + x + 1$ is irreducible over $\mathbb{Q}$. However, for odd primes $p$, the splitting field $\mathbb{Q}(\zeta_p)$ is equal to $\mathbb{Q}(\zeta_{2p})$. Thus, $\zeta_p = \zeta_{2p}^2$, that is, $\zeta_{p}$ has a square root in the same field. So $\zeta_{2p}$ has a minimal polynomial of degree $p-1$ and it is a root of $f_{p-1}(x)$, which has degree $2(p-1)$. So, $f_{p-1}(x)$ must split into two irreducible factors.
Therefore, the answer is yes, $f_n(x)$ is always reducible when $n\geq 2$.
