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: $$\begin{array}{rl}0& =\sum _{i=1}^n(a_i-2a_{i+1}{)}^2-\sum _{i=1}^n{a}_i^2\\ & =\sum _{i=1}^n{a}_i^2-4\sum _{i=1}^n{a}_i{a}_{i+1}+4\sum _{i=1}^n{a}_{i+1}^2-\sum _{i=1}^n{a}_i^2\\ & =2\left(\sum _{i=1}^n{a}_i^2-2\sum _{i=1}^n{a}_i{a}_{i+1}+\sum _{i=1}^n{a}_{i+1}^2\right)\\ & =2\sum _{i=1}^n(a_i-a_{i+1}{)}^2\end{array}$$ Hence, $a_1=a_2=\cdots =a_n$. The subsequent steps are the same as the main solution.