, 2013

First solution. Notice that we have
$$\begin{aligned} \left(a_{1}-n\right)^{2}+\cdots+\left(a_{n}-n\right)^{2} & =\left(a_{1}^{2}+\cdots+a_{n}^{2}\right)-2 n\left(a_{1}+\cdots+a_{n}\right)+n^{3} \\ & \leq n^{3}+1-2 n^{3}+n^{3}=1 . \end{aligned}$$
Therefore, there are two cases:

The first case is when $a_{1}=a_{2}=\cdots=a_{n}=n$. This is a solution since it satisfies both inequalities.

The second case is when there exists $1 \leq i_{0} \leq n$ such that $|a_{i_{0}}-n|=1$ and $a_{i}=n$ for all $1 \leq i \leq n$ with $i \neq i_{0}$.

If $a_{i_{0}}=n-1$, the first inequality becomes $n^{2}-1=a_{1}+\cdots+a_{n} \geq n^{2}$ which is impossible.

If $a_{i_{0}}=n+1$, the second inequality becomes $n^{3}+2 n+1=a_{1}^{2}+\cdots+a_{n}^{2} \leq n^{3}+1$ which is also impossible.

Hence, the only solution to both inequalities is given by $a_{1}=a_{2}=\cdots= a_{n}=n$.

Second solution. We have by Cauchy-Schwartz inequality
$$\left(n^{2}+1\right)^{2}>n^{4}+n \geq n\left(a_{1}^{2}+\cdots+a_{n}^{2}\right) \geq\left(a_{1}+\cdots+a_{n}\right)^{2} \geq n^{4} .$$
We deduce that
$$a_{1}+\cdots+a_{n}=n^{2}$$
and
$$n^{3}+1 \geq a_{1}^{2}+\cdots+a_{n}^{2} \geq n^{3}$$
and the right hand side equality occurs if and only if $a_{1}=\cdots=a_{n}=n$.

Now, assume, looking for a contradiction, that $a_{1}^{2}+\cdots+a_{n}^{2}=n^{3}+1$. Since a number and all his powers have the same parity, we deduce that
$$n \equiv n^{2} \equiv a_{1}+\cdots+a_{n} \equiv a_{1}^{2}+\cdots+a_{n}^{2} \equiv n^{3}+1 \equiv n+1 \quad \bmod 2$$
which is impossible.

Hence, the only solution to both inequalities is given by $a_{1}=a_{2}=\cdots= a_{n}=n$.