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

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    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)
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    Quote Originally Posted by elim View Post
    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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