In this year's team selection test to the international mathematical Olympiad in Japan, a interesting diophantine equation appeared as the final problem of the second day. 

> Find all quadruples $(a,b,c,d)$ of positive integers, such that $2^a3^b + 4^c5^d = 2^b3^a + 4^d5^c.$

It is clear that we must have $\min \{a,2c\}= \min \{b,2d\}$, and those quadruples satisfying $a = b$ and $c = d$ are trivial solutions. How could I proceed? And is there any generalization of this problem? It seems excessively hard so I decide to post it also in this forum. 

Any advice is welcomed. 

Also posted [here][1]. 


  [1]: https://math.stackexchange.com/posts/4949302