Japan Mathematical Olympiad

Let $a_{n+1} = a_1$, $a_0 = a_n$, and let $M$ and $m$ denote the maximum and minimum values of $a_1, a_2, \dots, a_n$, respectively. Remark that the maximum and minimum values of $a_1 - 2a_2, a_2 - 2a_3, \dots, a_n - 2a_1$ are also $M$ and $m$, respectively. Take $s$ satisfying $a_s = m$. Then, from $M \ge a_{s-1} - 2a_s \ge m - 2m$, we have $M + m \ge 0$. Therefore, if we choose $t$ satisfying $a_t = M$, then from $a_{t-1} - 2a_t \ge m$, it follows that $a_{t-1} \ge 2a_t + m = 2M + m \ge M$, implying $a_{t-1} = M$. Hence, by induction, we conclude that $a_i = M$ for any $i$. Consequently, from the assumption, we have $M = M - 2M$, which implies $M = 0$. Thus, it is necessary for $a_i = 0$ for any $i$. Conversely, the problem assumption is satisfied for this case. Therefore, the solution is $(a_1, a_2, \dots, a_n) = (0, 0, \dots, 0)$.

Another Solution. Let $a_{n+1} = a_1$, then from the assumption, we have:
$$\begin{align*} 0 &= \sum_{i=1}^{n} (a_i - 2a_{i+1})^2 - \sum_{i=1}^{n} a_i^2 \\ &= \sum_{i=1}^{n} a_i^2 - 4 \sum_{i=1}^{n} a_i a_{i+1} + 4 \sum_{i=1}^{n} a_{i+1}^2 - \sum_{i=1}^{n} a_i^2 \\ &= 2 \left( \sum_{i=1}^{n} a_i^2 - 2 \sum_{i=1}^{n} a_i a_{i+1} + \sum_{i=1}^{n} a_{i+1}^2 \right) \\ &= 2 \sum_{i=1}^{n} (a_i - a_{i+1})^2 \end{align*}$$
Hence, $a_1 = a_2 = \dots = a_n$. The subsequent steps are the same as the main solution.