Mathematica competitions in Croatia

Squaring the given equations we get $$\begin{array}{c}4x_k^2-4x_k{x}_{k+1}+x_{k+1}^2=x_{k+2}^2\text{for }k\in \{1,2,\dots ,98\},\\ 4x_{99}^2-4x_{99}{x}_{100}+x_{100}^2=x_1^2,4x_{100}^2-4x_{100}{x}_1+x_1^2=x_2^2.\end{array}$$ Adding these equations we get $$4(x_1^2+x_2^2+\cdots +x_{100}^2)-4(x_1{x}_2+\cdots +x_{99}{x}_{100}+x_{100}{x}_1)=0.$$ After regrouping and dividing by 2 we get $$(x_1-x_2{)}^2+\cdots +(x_{99}-x_{100}{)}^2+(x_{100}-x_1{)}^2=0.$$ Since all the terms are non-negative, they all have to be equal to zero, so we have $x_1=\cdots =x_{100}$.