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

**elim** · February 15th 2011, 06:16 PM

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

**Opalg** · February 16th 2011, 01:50 PM

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 $P_m(x^n)$ .

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

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