## Prove that a given family of polynomials is reducible over ZZ

Prove that $\psi_n(x) = \sum_{i=0}^n x^i$ is irreducible over $\mathbb{Z}$ if and only if $n+1$ is prime.

Note that if $n+1$ is prime, then $\psi_n$ is cyclotomic of prime index. We have already seen that $\psi_n \circ (x+1)$ is Eisenstein at $n+1$ and hence irreducible. (Note that $\psi_n(x) = (x^n - 1)/(x-1)$ and use the Binomial theorem.)

Now suppose $n+1 = ab$ is composite. Using the division algorithm, every number in $[0,n]$ can be written uniquely in the form $ia+j$ where $i \in [0,b-1]$ and $j \in [0,a-1]$. Thus we have $\psi_n(x) = \sum_{j=0}^{b-1} \sum_{i=0}^{a-1} x^{ia+j}$ $= \left( \sum_{i=0}^{b-1} x^{ia} \right) \left( \sum_{j=0}^{a-1} x^j \right)$, so that $\psi_n$ is reducible.