# Math Help - Find m,n ≥1 such that 1+x+…+x^m = P（x) | P(x^n)

1. ## Find m,n ≥1 such that 1+x+…+x^m = P（x) | P(x^n)

Find $(m,n)\; \left(m,n \in\mathbb{N}^+\right)$ such that $1+x+\cdots+x^m := P(x) | P(x^n)$

2. Originally Posted by elim
Find $(m,n)\; \left(m,n \in\mathbb{N}^+\right)$ such that $1+x+\cdots+x^m := P(x) | P(x^n)$
Let $P_m(x) = 1+x+\ldots+x^m$. Then $(1-x)P_m(x) = 1-x^{m+1}$. Next, $1-x^{m+1}$ divides $1-x^{n(m+1)} = (1-x^n)P_m(x^n)$. If m+1 and n are co-prime then none of the (complex) linear factors of $1-x^m$, apart from $1-x$, divides $1-x^n$, and therefore $P_m(x)$ divides $P_m(x^n)$. My guess is that gcd(m+1,n)=1 is a necessary and sufficient condition for $P_m(x)$ to divide $P_m(x^n)$.