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 April 15th, 2012, 05:14 AM #1 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Another logarithmic diophantine problem !!! How to prove that x^y = y^x (x is not = y) has no solution if y > 2 ????
 May 25th, 2012, 04:21 AM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Another logarithmic diophantine problem !!! I found a method for reducing it into normal diophantine form: x^y=y^x, assume x,y is very large (x,y integer) then y/ln(y) = x/ln(x) implies p(y)=p(x) where p is the prime counting function. is it possible to solve from here!?
 May 25th, 2012, 05:19 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,310 Thanks: 1981 As it's not true, there's no way to prove it.
 May 25th, 2012, 05:56 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,913 Thanks: 1113 Math Focus: Elementary mathematics and beyond Re: Another logarithmic diophantine problem !!! Can you provide a counterexample?
 May 25th, 2012, 06:42 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,310 Thanks: 1981 x = 2, y = 4 (there are infinitely many others if x and y needn't be integers).
May 25th, 2012, 11:17 PM   #6
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Re: Another logarithmic diophantine problem !!!

Quote:
 Originally Posted by skipjack x = 2, y = 4 (there are infinitely many others if x and y needn't be integers).
yes, this is the only case when x,y is integer and there are no other cases, I'm sure

 May 26th, 2012, 07:49 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,310 Thanks: 1981 Example of non-integer solution: x = 2^(2/3), y = 2^(8/3).
May 26th, 2012, 10:19 AM   #8
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Re: Another logarithmic diophantine problem !!!

Quote:
 Originally Posted by skipjack Example of non-integer solution: x = 2^(2/3), y = 2^(8/3).
Yes there are infinitely many solutions if x,y are non-integer, but I'm saying about integral x,y and what I said before, I'm pretty sure that there exist no solutions to the Diophantine equation: x^y=y^x except (x=2, y=4); (x=4,y=2) when x,y are distinct, greater than 1, integer;
now I'm asking you again, how to prove the above statement without the use of calculus?

 September 21st, 2012, 09:07 PM #9 Global Moderator   Joined: Dec 2006 Posts: 20,310 Thanks: 1981 Start by solving the equation, as explained in this post, then consider when the solutions found can be integers.
September 22nd, 2012, 08:02 AM   #10
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Re: Another logarithmic diophantine problem !!!

Quote:
 Originally Posted by skipjack Start by solving the equation, as explained in this post, then consider when the solutions found can be integers.
Yes, I know this can be solved using Lambert W Function and Calculus (there are several other approaches too). But actually I need a purely Diophantine approach.

Perhaps this would be appropriate if posted to the NT section. I posted this after a few months of joining so I wasn't familiar enough with this forum.

If moderators agree, then I could create another topic in the NT section asking about a Diophantine method to solve x^y = y^x where x,y are integers. Then mods can post a link to that topic and lock this topic.

I don't want it to be moved since it's already been months before I posted it and due to the discussion here, people can easily get confused about what I am exactly asking. Please inform me if any of you moderators agree with this.

Balarka

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