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!

I am a bit surprised that the answers given so far that invoke cyclotomic polynomials don't go all the way to exhibit a factorisation of $1+x^2+\cdots+x^{2n}$. Using $x^m-1=\prod_{d\mid m}\Phi_d(x)$ one gets
$$
 1+x^2+\cdots+x^{2n}=\frac{x^{2n+2}-1}{x^2-1}=\prod_{\substack{d\mid 2(n+1) \\ d\notin\{1,2\}}}\Phi_d(x),
$$
and since cyclotomic polynomials are irreducible over the rational numbers, this is precisely the factorisation of your polynomial in $\Bbb Q[x]$. The product is empty for $n=0$, has a single factor $\Phi_4(x)=x^2+1$ for $n=1$; in all other cases it has at least two factors, namely the polynomials $\Phi_{n+1}(x)$ and $\Phi_{2(n+1)}(x)$. So in particular it is always reducible for $n\geq2$.
