74th NMO Selection Tests for JBMO

Let $\tau (n)=k$ and let $d$ be a divisor of $n$, such that $d\le \sqrt{n}$. Since $\frac{n}{d}$ also divides $n$ and $\frac{n}{d}\ge \sqrt{n}$, we infer that $\tau (n)\le 2\sqrt{n}$, for any $n\in \mathbb{N}\setminus \{0\}$, (1).

Also, if $1=d_1<d_2<\cdots <d_k=n$ are the divisors of $n$, then $$2\sigma (n)=(d_1+d_k)+(d_2+d_{k-1})+\cdots +(d_k+d_1)=\left(d_1+\frac{n}{d_1}\right)+\cdots +\left(d_k+\frac{n}{d_k}\right).$$ But, if $d$ divides $n$, then $1+n-d-\frac{n}{d}=\frac{n(d-1)}{d}-(d-1)=\frac{(n-d)(d-1)}{d}\ge 0$. Consequently, $\sigma (n)\le \frac{1}{2}\tau (n)\cdot (n+1)$, (2).

We infer that $n=\sigma (\tau (n))\le \frac{1}{2}\tau (\tau (n))\cdot (\tau (n)+1)\le \frac{1}{2}\cdot 2\sqrt{\tau (n)}\cdot (2\sqrt{n}+1)\le \sqrt{2\sqrt{n}}\cdot (2\sqrt{n}+1)$.

Successively, this leads to $$n^2\le 2\sqrt{n}(4n+4\sqrt{n}+1)\iff n\sqrt{n}\le 2(4n+4\sqrt{n}+1)\iff \sqrt{n}\le 8+\frac{8}{\sqrt{n}}+\frac{2}{n}.$$ Suppose, by contradiction, that $n\ge 81$. Then $\sqrt{n}\ge 9$ and $8+\frac{8}{9}+\frac{2}{81}<9\le \sqrt{n}$, which is a contradiction. Hence, $n\le 80$, which means that $n$ can only have one of the following forms: $1,p,p^2,p^3,p^4,p^5,pq,p^{2}q,p^{3}q,p^{4}q,p^2{q}^2,p^3{q}^2,pqr,p^{2}qr$, where $p,q,r$ are prime numbers.

Every such possible $n$ has at most 12 divisors. Studying all cases for $\tau (n)\in \{1,2,\dots ,12\}$, we easily find that the only solutions are $n=1,n=3,n=4$ and $n=12$.