Croatian Mathematical Society Competitions

Let $a$ and $b$ be the lengths of legs, $c$ be the length of hypotenuse, and $r = 4$ be the radius of the incircle in the given triangle.
Expressing its area in two different ways, we get
$$r \cdot \frac{a + b + c}{2} = \frac{ab}{2},$$
$$ab - 4a - 4b = 4c.$$
Since $c^2 = a^2 + b^2$, it follows that $ab(ab - 8a - 8b + 32) = 0$. Considering that $a$ and $b$ are positive integers, we have $ab - 8a - 8b + 32 = 0$, i.e. $(a - 8)(b - 8) = 32$, hence both $a - 8$ and $b - 8$ must be divisors of $32$.
Without loss of generality, we can assume $a \leq b$, i.e. $a - 8 \leq b - 8$. Therefore, among
$$\begin{align*} a - 8 &= -32, & a - 8 &= 1, \\ a - 8 &= -16, & a - 8 &= 2, \\ a - 8 &= -8, & a - 8 &= 4, \end{align*}$$
we find that $(a, b) \in \{(9, 40), (10, 24), (12, 16)\}$ satisfy the conditions of the problem.
Finally, these are all solutions:
$$(a, b) \in \{(9, 40), (10, 24), (12, 16), (16, 12), (24, 10), (40, 9)\}.$$