3. Let $p$ and $q$ be given distinct prime numbers. Find all ordered pairs $(m, n)$ of natural numbers that satisfy the equation $p \cdot m + q \cdot n = m \cdot n$.
First method:

Let's transform the original equation:

$m \cdot n - p \cdot m - q \cdot n + p \cdot q = p \cdot q$,

$(m - q) \cdot (n - p) = p \cdot q$.

From this, it follows that $m - q$ and $n - p$ must have the same sign. Also, it is clear that $m \neq q$ and $n \neq p$, since $p \cdot q > 0$. From the original equation, it follows that it cannot be $m \cdot n + m \cdot n = 2 \cdot m \cdot n$, i.e., $m \cdot n < 0$, which contradicts the assumption that $m$ and $n$ are natural numbers.

Therefore, $m - q$ and $n - p$ must be positive*, and we have four possibilities:

$m - q = p q, \quad n - p = 1$,

$m - q = p, \quad n - p = q$,

$m - q = q, \quad n - p = p$,

$m - q = 1, \quad n - p = p q$.

From (1) we get the solution $(m, n) = (p q + q, p + 1)$,

from (2) we get the solution $(m, n) = (p + q, p + q)$,

from (3) we get the solution $(m, n) = (2 q, 2 p)$,

and from (4) we get the solution $(m, n) = (q + 1, p q + p)$.

*Note: The student does not have to immediately notice that they can eliminate the cases where $m - q$ and $n - p$ are negative. In that case, they will have eight instead of four cases, but all cases where $m - q$ and $n - p$ are negative will not result in a solution.

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