If $n$ is not prime, can $X^{n-1}+\dotsb+X+1$ be irreducible?

Question

If $n$ is not prime, can $X^{n-1}+\dotsb+X+1$ be irreducible over $\mathbb Q$ ? If $n$ is even, there is always $-1$ as solution, so it's necessarily reducible, but what happen iff $n$ is odd and not prime ?

Answer 1

The $n$-th cyclotomic polynomial is defined as $$ \Phi_n(X)=\prod_{\xi}(X-\xi) $$ where $\xi$ runs through the primitive $n$-th roots of unity. This is a polynomial with integer coefficients.

It follows that $$ X^n-1=\prod_{d\mid n}\Phi_d(X) $$ because any $n$-th root of unity is a primitive $d$-th root for a unique divisor $d$ of $n$ (because the roots of unity form a cyclic group of order $n$).

Thus $X^{n-1}+\dots+X+1$ is only irreducible when $n$ is prime.

Answer 2

If $n=pq$ then $$1+x+x^2+\cdots+x^{n-1}=\left(1+x+x^2+\cdots+x^{p-1}\right)\left(1+x^{p}+x^{2p}+\cdots+x^{(q-1)p}\right)$$

This can be seen if we write:

$$1+x+x^2+\cdots=x^{n-1}=\frac{x^n-1}{x-1}$$

But $x^{n}-1=(x^p)^q-1$ is divisible by $x^p-1$ and hence:

$$1+x+x^2+\cdots=x^{n-1}=\frac{x^n-1}{x^p-1} \frac{x^p-1}{x-1}$$