Baltic Way Problem Shortlist, 2011

For convenience we call $x_{2011}$ also $x_0$. Let $k$ be the largest of the numbers $x_1, \dots, x_{2011}$, and consider an equation $x_{n-1} + x_n = 2k$, where $1 \le n \le 2011$. Hence we get $2 \max(x_{n-1}, x_n) \ge x_{n-1} + x_n = 2k$, so either $x_{n-1}$ or $x_n$ is $\ge k$. Since $x_{n-1} \le k$, we then have $x_{n-1} = k$, and then also $x_n = 2k - x_{n-1} = 2k - k = k$. That is, in such an equation both variables on the left equal $k$. Now let $\mathcal{E}$ be the set of such equations, and let $\mathcal{S}$ be the set of subscripts on the left of these equations. From $x_n = k \forall n \in \mathcal{S}$ we get $|\mathcal{S}| \le |\mathcal{E}|$. On the other hand, since the total number of appearances of these subscripts is $2|\mathcal{E}|$ and each subscript appears on the left in no more than two equations, we have $2|\mathcal{E}| \le 2|\mathcal{S}|$. Thus $2|\mathcal{E}| = 2|\mathcal{S}|$, so for each $n \in \mathcal{S}$ the set $\mathcal{E}$ contains both equations with the subscript $n$ on the left. Now assume $1 \in \mathcal{S}$ without loss of generality. Then the equation $x_1 + x_2 = 2k$ belongs to $\mathcal{E}$, so $2 \in \mathcal{S}$. Continuing in this way we find that all subscripts belong to $\mathcal{S}$, so $x_1 = x_2 = \dots = x_{2011} = k$.

Again we call $x_{2011}$ also $x_0$. Taking the square on both sides of all the equations and adding the results, we get
$$\sum_{n=1}^{2011} (x_{n-1} + x_n)^2 = 4 \sum_{n=1}^{2011} x_n'^2 = 4 \sum_{n=1}^{2011} x_n^2,$$
which can be transformed with some algebra into
$$\sum_{n=1}^{2011} (x_{n-1} - x_n)^2 = 0.$$
Hence the assertion follows.