The polynomial $x^{{n-1}} + x^{n-2} + x^{{n-3}} +\dots + x +1$ is reducible when $n$ is composite

Question

Is $P(x)=(x^{{n-1}} + x^{n-2} + x^{{n-3}} +\dots + x +1)$ reducible if $n>1$ and $n$ not prime?

If $n-1$ is odd, $(x+1)|P(x)$, so if $n$ is even with $n>2$ I can write $$P(x)=(x+1) Q(x)$$ so is reducible.

If $n=2$, $P(x)=x+1$ is not reducible.

Some hint to help me conclude that, because $n=9$ is not prime and $x+1$ does not divide $P(x)$. If I use $(x-1)P(x)= x^{{n}} -1$, how do I conclude that $P(x)$ is reducible?

Answer 1

Multiply your expression by $x-1$, resulting in $x^n-1$. If n is composit, can you see simple factors different from $x-1$?

Answer 2

Hint: if $a\mid n$, show that $$ \left.\frac{x^a-1}{x-1}\,\middle|\,\frac{x^n-1}{x-1}\right. $$