## Problem 1

Determine all pairs of natural numbers $(n, m)$ that simultaneously satisfy the following conditions:

i) The number $n$ is composite;

ii) If the numbers $d_{1}, d_{2}, \ldots, d_{k}, k \in \mathbb{N}^{*}$ are all the proper divisors of $n$, then the numbers $d_{1}+1, d_{2}+1, \ldots, d_{k}+1$ are all the proper divisors of $m$.
Solution (Prof. Cristian Mangra, Bucharest):

Answer: $(n, m) \in\{(4,9),(8,15)\}$

For $k=1$, we get $n=p^{2}$, where $p$ is a prime number. We deduce that $m=(p+1)^{2}$, where $p+1$ is a prime number. We obtain $p=2, q=3$ and the pair $(n, m)=(4,9)$.

For $k \geq 2$, consider $d_{1}<d_{2}<\ldots<d_{k}$ all the proper divisors of $n$.

Then $n=d_{1} d_{k}=d_{2} d_{k-1}$ and $m=\left(d_{1}+1\right)\left(d_{k}+1\right)=\left(d_{2}+1\right)\left(d_{k-1}+1\right)$, from which we deduce that $d_{1}+d_{k}=d_{2}+d_{k-1}$, i.e., $d_{1}+\frac{n}{d_{1}}=d_{2}+\frac{n}{d_{2}}$, equivalent to $\left(d_{1}-d_{2}\right)\left(1-\frac{n}{d_{1} d_{2}}\right)=0$.

Therefore $n=d_{1} d_{2}$, so $k=2$, i.e.,
a) $n=d_{1} d_{2}$, where $d_{1}$ and $d_{2}$ are different prime numbers, so $m=\left(d_{1}+1\right)\left(d_{2}+1\right)$. Since $d_{1}+1$ and $d_{2}+1$ must be different primes, and $d_{2}+1$ is even, there is no solution. b) $n=d_{1}^{3}$, where $d_{1}$ is a prime number, i.e., $m=\left(d_{1}+1\right)\left(d_{1}^{2}+1\right)$ and $d_{1}+1, d_{1}^{2}+1$ are prime numbers. We obtain $d_{1}=2$ and the pair $(n, m)=(8,15)$.