Skip to main content
MathOverflow

Return to Question

Post Closed as "Not suitable for this site" by Will Jagy, Daniele Tampieri, Felipe Voloch, Daniel Weber, user44191
Post Undeleted by Lasting Howling
Post Deleted by Lasting Howling
deleted 49 characters in body
Source Link
Lasting Howling
Lasting Howling
  • 73
  • 4

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.

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.

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 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.

added 52 characters in body
Source Link
Lasting Howling
Lasting Howling
  • 73
  • 4

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 herehere.

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.

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.

Source Link
Lasting Howling
Lasting Howling
  • 73
  • 4

On the Diophantine equation $2^a3^b + 4^c5^d = 2^b3^a + 4^d5^c$

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.