MathDB
China TST 1999 sequence

Source: China TST 1999, problem 5

May 22, 2005
calculusintegrationalgebra unsolvedalgebra

Problem Statement

For a fixed natural number m≥2m \geq 2, prove that a.) There exists integers x1,x2,…,x2mx_1, x_2, \ldots, x_{2m} such that xixm+i=xi+1xm+i−1+1,i=1,2,…,m(∗)x_i x_{m + i} = x_{i + 1} x_{m + i - 1} + 1, i = 1, 2, \ldots, m \hspace{2cm}(*) b.) For any set of integers {x1,x2,…,x2m\lbrace x_1, x_2, \ldots, x_{2m} which fulfils (*), an integral sequence …,y−k,…,y−1,y0,y1,…,yk,…\ldots, y_{-k}, \ldots, y_{-1}, y_0, y_1, \ldots, y_k, \ldots can be constructed such that ykym+k=yk+1ym+k−1+1,k=0,±1,±2,…y_k y_{m + k} = y_{k + 1} y_{m + k - 1} + 1, k = 0, \pm 1, \pm 2, \ldots such that yi=xi,i=1,2,…,2my_i = x_i, i = 1, 2, \ldots, 2m.
Back to ProblemsView on AoPS