MathDB

9

Part of 2008 Mathcenter Contest

Problems(1)

p_i(x)+p_j(x^2)+p_k(x^4)=p_{100}(x)

Source: Mathcenter Contest / Oly - Thai Forum 2008 R3 p9 https://artofproblemsolving.com/community/c3196914_mathcenter_contest

11/9/2022
Set PP as a polynomial function by pn(x)=∑k=0n−1xkp_n(x)=\sum_{k=0}^{n-1} x^k. a) Prove that for m,n∈Nm,n\in N, when dividing pn(x)p_n(x) by pm(x)p_m(x), the remainder is pi(x),∀i=0,1,...,m−1.p_i(x),\forall i=0,1,...,m-1. b) Find all the positive integers i,j,ki,j,k that make pi(x)+pj(x2)+pk(x4)=p100(x).p_i(x)+p_j(x^2)+p_k(x^4)=p_{100}(x). (square1zoa)
polynomialalgebra