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!

There is a closed quotient form for the polynomial using the geometric series formula:
$$\sum_{k=0}^n x^{2k}=\frac{x^{2n+2}-1}{x^2-1}.$$
The theory of cyclotomic polynomials tells us how to reduce polynomials of the form $x^m-1$. In particular, the irreducible factors of $x^m-1$ are $\Phi_d(x)$ for each divisor $d\mid m$ (no repeated factors are present in this factorization). See references for $\Phi_n(x)$ for further details.
Thus,
$$\frac{x^{2n+2}-1}{x^2-1}=\prod_{\substack{d\mid (n+1) \\ d\ne1}}\Phi_d(x^2).$$
If $n+1$ is composite then the above factorization involves more than one cyclotomic polynomial hence $f_n(x)$ is reducible. On the other hand, if $n>1$ is odd we have
$$\frac{x^{2(n+1)}-1}{x^2-1}=\frac{x^n+1}{x+1}\frac{x^n-1}{x-1}.$$
Both quotients are in fact integer-coefficient polynomials:
$$\frac{x^n-1}{x-1}=x^{n-1}+\cdots+x+1, \quad \frac{x^n+1}{x+1}=\frac{(-x)^n-1}{(-x)-1}=(-x)^{n-1}+\cdots+(-x)+1.$$
Note the latter depended on $n$ being odd so that $(-x)^n=-x^n$.
The only exception is $n=1$, and in that particular case we know that $x^2+1$ is irreducible.
