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!
 A: Note that  $f_n(x)(x^2-1) = x^{2n+2} - 1 = (x^{n+1}-1)(x^{n+1}+1)$. Now use unique factorization...  
A: 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.
A: 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 $n\geq 2$.
A: 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 $n\geq2$.
A: For odd $n$ you can always write:
$$f_n(x) = \sum_{i=0}^n x^{2i} = (x^2+1)(x^{2n-2}+x^{2n-6}+x^{2n-10}+...)$$
For even $n$ this trick no longer works.
