mersenneforum.org  

Go Back   mersenneforum.org > Great Internet Mersenne Prime Search > Math

Reply
 
Thread Tools
Old 2007-05-11, 23:42   #1
Citrix
 
Citrix's Avatar
 
Jun 2003

32·52·7 Posts
Default Modified fermat's last theorem

I found this problem on a website (lost link to the website).

The problem asks for solutions of x^x*y^y=z^z
Are any solutions possible?

If possible I think x and y have to be smooth. Any thoughts on how to approach this?


Last fiddled with by Citrix on 2007-05-11 at 23:43
Citrix is online now   Reply With Quote
Old 2007-05-12, 03:34   #2
wblipp
 
wblipp's Avatar
 
"William"
May 2003
New Haven

93816 Posts
Default

Quote:
Originally Posted by Citrix View Post
The problem asks for solutions of x^x*y^y=z^z
Are any solutions possible?
I suppose you want to exclude x=1 and the symmetric y=1?
wblipp is offline   Reply With Quote
Old 2007-05-12, 15:05   #3
Citrix
 
Citrix's Avatar
 
Jun 2003

110001001112 Posts
Default

yes
Citrix is online now   Reply With Quote
Old 2007-05-15, 06:58   #4
nibble4bits
 
nibble4bits's Avatar
 
Nov 2005

2×7×13 Posts
Default

It almost has to involve logorythms but I may be wrong in this intuition. :LOL: Likely!

Did you mean (x^x)*(y^y)=z^z?
Or x^(x(y^y))=z^z?

Last fiddled with by nibble4bits on 2007-05-15 at 06:59
nibble4bits is offline   Reply With Quote
Old 2007-05-15, 16:56   #5
victor
 
victor's Avatar
 
Oct 2005
Fribourg, Switzerlan

22×32×7 Posts
Default

Quote:
Originally Posted by nibble4bits View Post
It almost has to involve logorythms but I may be wrong in this intuition. :LOL: Likely!

Did you mean (x^x)*(y^y)=z^z?
Or x^(x(y^y))=z^z?
He meant x^x*y^y=z^z
And this means (x^x)*(y^y)=z^z

I think we can conjecture there is no integer solution.

btw, I tested x^x \cdot y^y=z^z for each x, y, z \in [2; 500]
victor is offline   Reply With Quote
Old 2007-05-15, 21:31   #6
R.D. Silverman
 
R.D. Silverman's Avatar
 
Nov 2003

1D2416 Posts
Default

Quote:
Originally Posted by victor View Post
He meant x^x*y^y=z^z
And this means (x^x)*(y^y)=z^z

I think we can conjecture there is no integer solution.

btw, I tested x^x \cdot y^y=z^z for each x, y, z \in [2; 500]
(1) This conjecture can not be considered as even remotely related to FLT.
FLT is a conjecture about algebraic (abelian) varieties. This is not one.

(2) Proving that this equation has no solution is easy. Hint: if a prime
divides the right side count how many times that it does. It must
divide the left side the same number of time.

This leads to a simple additive relation between x,y, and z. Which then
leads to the impossibility.
R.D. Silverman is offline   Reply With Quote
Old 2007-05-15, 22:11   #7
victor
 
victor's Avatar
 
Oct 2005
Fribourg, Switzerlan

22×32×7 Posts
Default

1) I didn't said it was related to FLT

2) Thanks a lot for these specifications, really interresting
victor is offline   Reply With Quote
Old 2007-05-16, 04:53   #8
Citrix
 
Citrix's Avatar
 
Jun 2003

62716 Posts
Default

Doesn't make a difference if it is related to FLT or not.

lets say for a given prime p, such that x=p^l*a and y=p^m*b and z=p^n*c we have

l*x+m*y=n*z

How does one proceed from here?
Citrix is online now   Reply With Quote
Old 2007-05-16, 09:31   #9
akruppa
 
akruppa's Avatar
 
"Nancy"
Aug 2002
Alexandria

246710 Posts
Default

Quote:
Originally Posted by victor View Post
1) I didn't said it was related to FLT
No, but Citrix originally did. See thread title.

Alex
akruppa is offline   Reply With Quote
Old 2007-05-16, 09:34   #10
R.D. Silverman
 
R.D. Silverman's Avatar
 
Nov 2003

22×5×373 Posts
Default

Quote:
Originally Posted by Citrix View Post
Doesn't make a difference if it is related to FLT or not.

lets say for a given prime p, such that x=p^l*a and y=p^m*b and z=p^n*c we have

l*x+m*y=n*z

How does one proceed from here?

Select a prime that only divides z to the first power.
R.D. Silverman is offline   Reply With Quote
Old 2007-05-16, 09:51   #11
Citrix
 
Citrix's Avatar
 
Jun 2003

157510 Posts
Default

Quote:
Originally Posted by R.D. Silverman View Post
Select a prime that only divides z to the first power.
How does this help? Sorry, I can't see the solution. If a prime divides Z, it must be a factor of x or y? So if n>0 then m or l must be >0 or =n.

Alex: the website called it modified LFT (not me)
Citrix is online now   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Conjecture pertaining to modified Fermat's theorem devarajkandadai Number Theory Discussion Group 12 2017-12-25 05:43
Modified Fermat's theorem devarajkandadai Number Theory Discussion Group 14 2017-11-12 20:04
modified Euler's generalisation of Fermat's theorem devarajkandadai Number Theory Discussion Group 1 2017-07-07 13:56
Modified Fermat pseudoprime devarajkandadai Number Theory Discussion Group 0 2017-06-24 12:11
Modified Fermat's theorem devarajkandadai Number Theory Discussion Group 2 2017-06-23 04:39

All times are UTC. The time now is 07:19.

Wed Dec 2 07:19:25 UTC 2020 up 83 days, 4:30, 1 user, load averages: 1.10, 1.33, 1.42

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.