China Western Mathematical Olympiad

Suppose $(a_1,a_2,\dots ,a_n)$ is an integer group which satisfies the conditions. Then by Cauchy's inequality we have $$a_1^2+\cdots +a_n^2\ge \frac{1}{n}(a_1+\cdots +a_n{)}^2\ge n^3.\stackrel{◯}{1}$$ Combining $a_1^2+\cdots +a_n^2\le n^3+1$, we see that it can only be $a_1^2+\cdots +a_n^2=n^3$ or $a_1^2+\cdots +a_n^2=n^3+1$.

If it is the former, then by the condition for Cauchy's inequality to take the equality sign, we can obtain $a_1=\cdots =a_n$. This requires $a_1^2=n^2$, $1\le i\le n$. Combining $a_1+\cdots +a_n\ge n^2$, we have $a_1=\cdots =a_n=n$.

If it is the latter, then let $b_i=a_i-n$, then we have $$\begin{gathered} b_1^2+b_2^2+\cdots +b_n^2=\sum _{i=1}^n{a}_i^2-2n\sum _{i=1}^n{a}_i+n^3 \\ =2n^3+1-2n\sum _{i=1}^n{a}_i\le 1. \end{gathered}$$ Thus $b_1^2$ can only be $0$ or $1$, and there is at the most one among $b_1^2$, $b_2^2$, ..., $b_n^2$ to be $1$. If $b_1^2$, ..., $b_n^2$ all are zero, then $a_i=n$, ${\sum }_{i=1}^n{a}_i^2=n^3\ne n^3+1$. It leads to a contradiction. If there is just one among $b_1^2$, ..., $b_n^2$ to be $1$, then ${\sum }_{i=1}^n{a}_i^2=n^3\pm 2n+1\ne n^3+1$. Again, it leads to a contradiction.

Consequently, we obtain that it can only be $(a_1,\dots ,a_n)=(n,\dots ,n)$.