24th Balkan Mathematical Olympiad

Let for some $n\in {\mathbb{N}}^{\ast }$ we have $$\sqrt{\sigma (1)+\sqrt{\sigma (2)}+\sqrt{\cdots +\sqrt{\sigma (n)}}}=r_1\in \mathbb{Q}.(1)$$ Squaring the two parts of (1) we conclude that the number $\sqrt{\sigma (2)+\sqrt{\sigma (3)}+\sqrt{\cdots +\sqrt{\sigma (n)}}}$ is rational. Similarly we obtain that, for each $k\in \{1,2,\dots ,n\}$, the number $$r_k:=\sqrt{\sigma (k)+\sqrt{\sigma (k+1)}+\sqrt{\cdots +\sqrt{\sigma (n)}}}\in \mathbb{Q}.$$ In the sequel we denote by $a_k:=\sqrt{n+\sqrt{n+\sqrt{\cdots +\sqrt{n}}}}$, for each $k\in {\mathbb{N}}^{\ast }$. Using induction we can prove that $a_k<\sqrt{n+1}$, for every $k\in {\mathbb{N}}^{\ast }$, and therefore $$r_1=\sqrt{\sigma (1)+\sqrt{\sigma (2)}+\sqrt{\cdots +\sqrt{\sigma (n)}}}<a_n<\sqrt{n+1}.$$ If $\ell$ is a positive integer satisfying the inequalities ${\ell }^2\le n<(\ell +1{)}^2$, then for some $i\in \{1,2,\dots ,n\}$ we have $\sigma (i)={\ell }^2$. Now we distinguish two cases

1ª case: $i\ne n$ Then we have $$\ell <\sqrt{{\ell }^2+\sqrt{\sigma (i+1)}+\sqrt{\cdots +\sqrt{\sigma (n)}}}<\sqrt{n+1}<\ell +2,$$ from which it follows that $$\sqrt{{\ell }^2+\sqrt{\sigma (i+1)}+\sqrt{\cdots +\sqrt{\sigma (n)}}}=\ell +1.(2)$$ Squaring both parts of (2) we get $$2\ell +1=\sqrt{\sigma (i+1)+\sqrt{\cdots +\sqrt{\sigma (n)}}}<\sqrt{n+1}<\ell +2,$$ from which it follows that $\ell <1$, which is absurd.

2ª case: $i=n$ If $\ell >1$, then ${\ell }^2-1\in \{\sigma (1),\dots ,\sigma (n-1)\}$. Let now $j<n$ such that $\sigma (j)={\ell }^2-1$. Similarly, as in the first case, we get $$\begin{array}{r}\ell <\sqrt{{\ell }^2-1+\sqrt{\sigma (j+1)+\sqrt{\cdots +\sqrt{{\ell }^2}}}}<\sqrt{n+1}<\ell +2\\ \Rightarrow \sqrt{{\ell }^2-1+\sqrt{\sigma (j+1)+\sqrt{\cdots +\sqrt{{\ell }^2}}}}=\ell +1\\ \Rightarrow 2\ell +2=\sqrt{\sigma (j+1)+\sqrt{\cdots +\sqrt{{\ell }^2}}}<\sqrt{n+1}<\ell +2,\end{array}$$ which is absurd. If $\ell =1$, then $n\in \{1,2,3\}$. Checking all the cases it is easy to see that only for $n=1$ and $n=3$ there exist permutations satisfying the relation we seek. More explicitly, for $n=1$ we have $\sqrt{1}=1$, while for $n=3$ we have: $$\sqrt{2+\sqrt{3}+\sqrt{1}}=2.\text{ For }n=2\text{ does not exist such a permutation.}$$