, 2019

Solution:

This is equivalent to the existence of nonnegative integers $c$ and $d$ such that $3b - 1 = c(2a + 1)$ and $3a - 1 = d(2b + 1)$. Then
$$c d = \frac{(3b - 1)(3a - 1)}{(2a + 1)(2b + 1)} = \frac{3a - 1}{2a + 1} \cdot \frac{3b - 1}{2b + 1} < \frac{3}{2} \cdot \frac{3}{2} = 2.25.$$
Neither $c$ nor $d$ can equal $0$ since that would give $a = \frac{1}{3}$ or $b = \frac{1}{3}$, so $cd \leq 2.25$ implies $(c, d) \in \{(1, 1), (2, 1), (1, 2)\}$. Substituting $(c, d)$ back in gives three systems of equations and the three solutions: $(2, 2), (12, 17), (17, 12)$.