lambruscokid wrote: We have $ 150$ numbers $ x_1,x_2, \cdots , x_{150}$ each of which is either $ \sqrt 2 + 1$ or $ \sqrt 2 - 1$ We calculate the following sum: $ S = x_1x_2 + x_3x_4 + x_5x_6 + \cdots + x_{149}x_{150}$ Can we choose the $ 150$ numbers such that $ S = 121$? And what about $ S = 111$? Notice that in sum there are $ 75$ terms each one can have its value independent of other. Moreover each can be $ (\sqrt 2 + 1)(\sqrt 2 - 1) = 1,(\sqrt 2 + 1)^{2} = (3 + 2\sqrt 2),$ or $ (\sqrt 2 - 1)^{2} = (3 - 2\sqrt 2)$ Consider $ x$ of them equal to $ 1, y$ of them equal to $ (3 + 2\sqrt 2)$ and $ z$ of them equal to $ (3 - 2\sqrt 2)$ Now $ x + y + z = 75$ And $ x + (3 + 2\sqrt 2)y + (3 - 2\sqrt 2)z = S$ as $ S$ is radical free $ \implies y = z \implies x$ is odd. Let $ x = 2n + 1$ $ \implies 2n + 1 + 2y = 75\implies n + y = 37$ (1) Also $ 2n + 1 + 6y = S$ Consider $ S = 121\implies n + 3y = 60$- (2) Now see (1) and (2) has no solution in integers. For $ S = 111\implies n + 3y = 55$- (3) Now from (1) and (3) we get $ y = 9, n = 28$ Hence for $ S = 111$ we have such set.

S=111 Solution! $x_{1} , x_{2} , x_{3} ... x_{18} = \sqrt 2+1$ $x_{19} , x_{20} , x_{21} ... x_{36} = \sqrt 2-1$ $x_{37} , x_{39} , x_{41} ... x_{149} = \sqrt 2+1$ $x_{38} , x_{40} , x_{42} ... x_{150} = \sqrt 2-1$

2017 Junior MO, Poland, second round P4 was the same question, but equation was changed to $$x_1x_2+x_2x_3+...+x_{98}x_{99}+x_{99}x_1=199$$