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

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

Originally Posted by elim

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

Let $\displaystyle P_m(x) = 1+x+\ldots+x^m$. Then $\displaystyle (1-x)P_m(x) = 1-x^{m+1}$. Next, $\displaystyle 1-x^{m+1}$ divides $\displaystyle 1-x^{n(m+1)} = (1-x^n)P_m(x^n)$. If m+1 and n are co-prime then none of the (complex) linear factors of $\displaystyle 1-x^m$, apart from $\displaystyle 1-x$, divides $\displaystyle 1-x^n$, and therefore $\displaystyle P_m(x)$ divides $\displaystyle P_m(x^n)$. My guess is that gcd(m+1,n)=1 is a necessary and sufficient condition for $\displaystyle P_m(x)$ to divide $\displaystyle P_m(x^n)$.