Japan Mathematical Olympiad

29900 Let $x_0=0$. Then, the given identity can be rewritten as: $$\begin{array}{rl}& x_0^2+x_1^2+\cdots +x_{25}^2=2+x_0{x}_1+x_1{x}_2+\cdots +x_{24}{x}_{25}+x_{25}{x}_0\\ & ⟺(x_0-x_1{)}^2+(x_1-x_2{)}^2+\cdots +(x_{24}-x_{25}{)}^2+(x_{25}-x_0{)}^2=4.\end{array}$$

26 integers $x_0-x_1,x_1-x_2,\dots ,x_{24}-x_{25},x_{25}-x_0$ add up to 0, and the sum of their squares equals 4. Therefore, each of these 26 numbers can only take value in $0,\pm 1,\pm 2$. If there is one of these 26 numbers taking value $\pm 2$, then all other numbers must be 0, but then the sum of all the 26 numbers cannot be equal to 0, contrary to our assumption. Therefore, each of these 26 numbers must take 0 or $\pm 1$. Since the sum of all these numbers is 0 and the sum of all the squares is 4, we can conclude that there are exactly two 1's and two -1's among them and all others must be 0. Furthermore, since $x_0=0$ and $x_1,x_2,\dots ,x_{25}$ are non-negative integers, we see that among the 26 numbers $x_0-x_1,x_1-x_2,\dots ,x_{24}-x_{25},x_{25}-x_0$ the numbers 1 and -1 can appear only in the order of -1, -1, 1, 1 or -1, 1, -1, 1. If we put these two 1's and two -1's among 26 possible spots keeping one of the above two orders, then we end up with a 25-tuple $(x_1,x_2,\dots ,x_{25})$ satisfying the requirement of the problem. For each of the two orders for the appearance of 1 and -1, there are $2\cdot {}_{26}{C}_4$ ways of choosing 4 spots for $\pm 1$ to go in, and therefore, there are $2\cdot {}_{26}{C}_4=29900$ 25-tuples satisfying the condition of the problem.