Subject (1). Determine the natural number $a=\frac{p+q}{r}+\frac{q+r}{p}+\frac{r+p}{q}$, where $p, q$, and $r$ are positive prime numbers.

## LUCIAN PETRESCU
Solution. If $p=q=r$ then $a=6$. If $p=q \neq r$ then we must have $p r \mid 2\left(p^{2}+r^{2}\right)$, so $p \mid 2 r^{2}$ and $r \mid 2 p^{2}$, which is impossible. If $p, q, r$ are distinct then we must have $p q r \mid p q(p+q)+q r(q+r)+r p(r+p)$, so $p \mid q r(q+r)$, $q \mid r p(r+p)$, $r \mid p q(p+q)$, which means $p \mid q+r$, $q \mid r+p$, $r \mid p+q$. Therefore, $p q r \mid p+q+r$, from which $p q r \leq p+q+r$, i.e., $\frac{3}{4}>\frac{1}{p q}+\frac{1}{q r}+\frac{1}{r p} \geq 1$, which is impossible.

Remark. One primary flaw is the use of the articulated word number, which already suggests that the answer is unique. Moreover, the manipulations to arrive at this answer are so trivial that I wouldn't see them required even in a class extemporal. I don't understand the results obtained in the competition at all...[^0]