Selection tests for the International Mathematical Olympiad 2013

First solution. Notice that we have $$\begin{array}{rl}{\left(a_1-n\right)}^2+\cdots +{\left(a_n-n\right)}^2& =\left(a_1^2+\cdots +a_n^2\right)-2n\left(a_1+\cdots +a_n\right)+n^3\\ & \le n^3+1-2n^3+n^3=1.\end{array}$$ 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 $1\le i_0\le n$ such that $|a_{i_0}-n|=1$ and $a_i=n$ for all $1\le i\le n$ with $i\ne i_0$.

If $a_{i_0}=n-1$, the first inequality becomes $n^2-1=a_1+\cdots +a_n\ge n^2$ which is impossible.

If $a_{i_0}=n+1$, the second inequality becomes $n^3+2n+1=a_1^2+\cdots +a_n^2\le n^3+1$ 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 $${\left(n^2+1\right)}^2>n^4+n\ge n\left(a_1^2+\cdots +a_n^2\right)\ge {\left(a_1+\cdots +a_n\right)}^2\ge n^4.$$ We deduce that $$a_1+\cdots +a_n=n^2$$ and $$n^3+1\ge a_1^2+\cdots +a_n^2\ge n^3$$ 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 $$n\equiv n^2\equiv a_1+\cdots +a_n\equiv a_1^2+\cdots +a_n^2\equiv n^3+1\equiv n+1\mathrm{mod}2$$ which is impossible.

Hence, the only solution to both inequalities is given by $a_1=a_2=\cdots =a_n=n$.