3. Find all ordered triples ( $m, n, p$ ) of positive rational numbers such that the numbers

$$
m+\frac{1}{n p}, n+\frac{1}{p m}, p+\frac{1}{m n}
$$

are all integers.

(Ru.munija)
3. Let $a=m n p$. The numbers $\frac{a+1}{m n}, \frac{a+1}{n p}$, and $\frac{a+1}{p m}$ are integers, so their product $\frac{(a+1)^{3}}{a^{2}}=k$ is also an integer. Therefore, $a$ is a root of the polynomial $x^{3}+(3-k) x^{2}+3 x+1=0$. From this, it follows that if $a=\frac{q}{r}(q, r \in \mathbb{N},(q, r)=1)$, then $q, r \mid 1$, hence $a=1$.

Therefore, $\frac{a+1}{m n}=2 p$ and similarly $2 m$ and $2 n$ are integers, so since $2 m \cdot 2 n \cdot 2 p=8$, the only solutions $(m, n, p)$ are the triples $(1,1,1),\left(2,1, \frac{1}{2}\right)$ and ( $4, \frac{1}{2}, \frac{1}{2}$ ) with their permutations.