Well, $a=3, b=1$ works.

That is only solution. This equation is equivalent to: $$2(2^{a-1}-1)-5(5^{b-1}-1)=0$$so it's easy to see that $4\nmid2(2^{a-1}-1)$ (else $a=1$ so $2-5^b=3$ and it is not true) so $2|5(5^{b-1}-1)$ but $4\nmid 5(5^{b-1}-1)$ or $5(5^{b-1}-1)=0$. Lets assume that $5(5^{b-1}-1)\neq 0 \Longrightarrow b\neq 1$, then $4\nmid 5(5^{b-1}-1)$ but $5(5^{b-1}-1)=5((5-1)(5^{b-2}+5^{b-3}+...+1))$ so its obviously not true. So because $b=1$ we get $a=3$. $\Box$

Flow25 wrote: That is only solution. This equation is equivalent to: $$2(2^{a-1}-1)-5(5^{b-1}-1)=0$$so it's easy to see that $4\nmid2(2^{a-1}-1)$ (else $a=1$ so $2-5^b=3$ and it is not true) so $2|5(5^{b-1}-1)$ but $4\nmid 5(5^{b-1}-1)$ or $5(5^{b-1}-1)=0$. Lets assume that $5(5^{b-1}-1)\neq 0 \Longrightarrow b\neq 1$, then $4\nmid 5(5^{b-1}-1)$ but $5(5^{b-1}-1)=5((5-1)(5^{b-2}+5^{b-3}+...+1))$ so its obviously not true. So because $b=1$ we get $a=3$. $\Box$ That's not correct, because $a=7, b=3$ is also a solution.

rafaello wrote: Flow25 wrote: That is only solution. This equation is equivalent to: $$2(2^{a-1}-1)-5(5^{b-1}-1)=0$$so it's easy to see that $4\nmid2(2^{a-1}-1)$ (else $a=1$ so $2-5^b=3$ and it is not true) so $2|5(5^{b-1}-1)$ but $4\nmid 5(5^{b-1}-1)$ or $5(5^{b-1}-1)=0$. Lets assume that $5(5^{b-1}-1)\neq 0 \Longrightarrow b\neq 1$, then $4\nmid 5(5^{b-1}-1)$ but $5(5^{b-1}-1)=5((5-1)(5^{b-2}+5^{b-3}+...+1))$ so its obviously not true. So because $b=1$ we get $a=3$. $\Box$ That's not correct, because $a=7, b=3$ is also a solution. Yes I think $a=3,b=1$ and $a=7, b=3$ are the only solutions. I just don't know how to prove it...

This looks easy, but is hard...

rafaello wrote: Flow25 wrote: That is only solution. This equation is equivalent to: $$2(2^{a-1}-1)-5(5^{b-1}-1)=0$$so it's easy to see that $4\nmid2(2^{a-1}-1)$ (else $a=1$ so $2-5^b=3$ and it is not true) so $2|5(5^{b-1}-1)$ but $4\nmid 5(5^{b-1}-1)$ or $5(5^{b-1}-1)=0$. Lets assume that $5(5^{b-1}-1)\neq 0 \Longrightarrow b\neq 1$, then $4\nmid 5(5^{b-1}-1)$ but $5(5^{b-1}-1)=5((5-1)(5^{b-2}+5^{b-3}+...+1))$ so its obviously not true. So because $b=1$ we get $a=3$. $\Box$ That's not correct, because $a=7, b=3$ is also a solution. true sorry i made mistake with adding at very beginning

There are at least three different methods to solve this equation in positive integers. But I agree that the problem is very amazing. I suggest one of possible ways: putting $a=m+7$ and $b=n+3$, rewrite the equation in the form $$5^3(5^n-1)=2^7(2^m-1).$$

https://artofproblemsolving.com/community/c6h27493

$Mod1024$ and $mod257$ This is also in Serbia TST 2019