66th 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 $$2^7(2^{\alpha }-1)=5^3(5^{\beta }-1).(1)$$ 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 $5^{\beta }\equiv 1(\mathrm{mod}{2}^7)$. One can easily deduce that $\beta \supseteq 32$.

2. Also, $A\supseteq 5^3$, so $2^{\alpha }\equiv 1(\mathrm{mod}125)$ 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 $2^{\alpha }\equiv 1(\mathrm{mod}97)$. 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).