Let $p$ be a prime number. How can we show that the polynomial $1+X+⋯+X^{p−1}$ cannot be written in the form $P\times Q$, where $P,Q\in \mathbb R[X]$ are non-constant polynomials with non-negative coefficients?
I know that this polynomial is irreducible in $\mathbb Q[X]$. If $p$ is not a prime number, then this polynomial can be written in that form. Suppose $p=mn$. Then $1+X+...+X^{p−1}$ $=(1+X+⋯+X^{m-1})$ $(1+X^m+X^{2m}+⋯+X^{(n-1)m}).$
Towards a contradiction, suppose $$1+x\cdots+x^{p−1}=P(x)\times Q(x)$$ where $P(x)= a_0 + a_1x+a_2x^2+\cdots + a_{r-1}x^{r-1} +x^r$, $\ Q(x)=b_0 + b_1x+\cdots + x^s$, $r,s\ge 1$ are integers and $a_0,a_1,\cdots, a_{r-1}, b_0,b_1,\cdots, b_{s-1}$ are real nonnegative numbers.
Thanks to this post, we know that each $a_i$ and each $b_j$ is either $0$ or $1$ and that $ a_0=b_0=1$.
Let $x=1$, we get $$p=P(1)\times Q(1).$$
$P(1)=1+\cdots+1$ is an integer $>1$. So is $Q(1)$. Hence, $p$ is not a prime number. That is a contradiction. Our proof is complete.