1. Determine all triples of positive integers $k, l$, and $m$ for which

$$
\frac{3 l+1}{3 k l+k+3}=\frac{l m+1}{5 l m+m+5} .
$$
1. First, we will invert the fractions in the given equation (both numerators contain positive numbers) and partially divide:

$$
\begin{aligned}
\frac{3 k l + k + 3}{3 l + 1} & = \frac{5 l m + m + 5}{l m + 1}, \\
\frac{k(3 l + 1) + 3}{3 l + 1} & = \frac{5(l m + 1) + m}{l m + 1}, \\
k + \frac{3}{3 l + 1} & = 5 + \frac{m}{l m + 1}.
\end{aligned}
$$

Since for natural numbers $l$ and $m$ it holds that $3 < 3 l + 1$ and $m < l m + 1$, the values of both fractions in the last equation lie in the interval $(0,1)$. We thus obtain

$$
k = 5 \quad \text{and consequently} \quad \frac{3}{3 l + 1} = \frac{m}{l m + 1}.
$$

From the second equation, after expanding, we get $3 l m + 3 = 3 l m + m$, hence $m = 3$, while $l$ can be any value.

The solutions to the problem are all triples $(k, l, m) = (5, l, 3)$, where $l$ is any natural number.

For a complete solution, award 6 points. For deriving $k = 5$, award 3 points, for $m = 3$ award 2 points, and for determining that the value of $l$ can be any, award 1 point. If the solver guesses only a finite number of solutions $(k, l, m)$, such as just one solution $(5, 1, 3)$, award 1 point. If the solver guesses and verifies the solution $(5, l, 3)$ for any natural number $l$, award an additional point.