MathDB

6

Part of 1994 Austrian-Polish Competition

Problems(1)

(x_2 - x_1)^2 + 2(x_2 +x_1) + 1 = n^2 , system of equations

Source: Austrian - Polish 1994 APMC

5/3/2020
Let n>1n > 1 be an odd positive integer. Assume that positive integers x1,x2,...,xn≥0x_1, x_2,..., x_n \ge 0 satisfy: {(x2−x1)2+2(x2+x1)+1=n2(x3−x2)2+2(x3+x2)+1=n2...(x1−xn)2+2(x1+xn)+1=n2\begin{cases} (x_2 - x_1)^2 + 2(x_2 +x_1) + 1 = n^2 \\ (x_3 -x_2)^2 + 2(x_3 +x_2) + 1 = n^2 \\ ...\\ (x_1 - x_n)^2 + 2(x_1 + x_n)+ 1 = n^2 \end {cases} Show that there exists j,1≤j≤nj, 1 \le j \le n, such that xj=xj+1x_j = x_{j+1}. Here xn+1=x1x_{n+1} = x_1.
system of equationsalgebra