Belarusian Mathematical Olympiad

Answer: $(a; b) = (3; 1)$ or $(7; 3)$.

(Solution by Y. Dubovik.) Note that if $a \le 7$ or $b \le 3$, then it is easy to verify that the only solutions are $(a; b) = (3; 1)$ and $(a; b) = (7; 3)$.

Now it remains to prove that there are no solutions with $a > 7, b > 3$. In this case we can write $a = 7 + \alpha, b = 3 + \beta$, where $\alpha, \beta \in \mathbb{N}$. Then the given equation $2^{7+\alpha} - 5^{3+\beta} = 3$ transforms to
$$2^7(2^\alpha - 1) = 5^3(5^\beta - 1). \qquad (1)$$
Suppose that (1) has a solution $(\alpha; \beta)$ in positive integers. Set $A = 2^7(2^\alpha - 1) = 5^3(5^\beta - 1)$.

1. From (1) it follows that $A \supseteq 2^7$, so $5^\beta \equiv 1 \pmod{2^7}$. One can easily deduce that $\beta \supseteq 32$.

2. Also, $A \supseteq 5^3$, so $2^\alpha \equiv 1 \pmod{125}$ and we deduce that $\alpha \supseteq 100$. In particular, $A \supseteq (2^{100} - 1)$, and so $A \supseteq (2^5 - 1) = 31$. It follows that $5^\beta - 1 \supseteq 31$, thus $\beta \supseteq 3$.

3. Thus we have $\beta \supseteq 96$, whence $A \supseteq (5^{96} - 1)$, so $A \supseteq 97$, which implies $2^\alpha \equiv 1 \pmod{97}$. Therefore, $\alpha \supseteq 48$. In particular, $A \supseteq (2^{48} - 1)$, then $A \supseteq (2^{16} - 1)$ and $A \supseteq (2^8 + 1) = 257$.

4. Finally, $5^\beta - 1 \supseteq 257$, which gives $\beta \supseteq 256$, whence $A \supseteq 2^8$, contrary to (1).