20070511, 23:42  #1 
Jun 2003
1,579 Posts 
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 20070511 at 23:43 
20070512, 03:34  #2 
"William"
May 2003
New Haven
3×787 Posts 

20070512, 15:05  #3 
Jun 2003
1,579 Posts 
yes

20070515, 06:58  #4 
Nov 2005
B6_{16} Posts 
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 20070515 at 06:59 
20070515, 16:56  #5  
Oct 2005
Fribourg, Switzerlan
2^{2}·3^{2}·7 Posts 
Quote:
And this means (x^x)*(y^y)=z^z I think we can conjecture there is no integer solution. btw, I tested for each 

20070515, 21:31  #6  
Nov 2003
2^{2}·5·373 Posts 
Quote:
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. 

20070515, 22:11  #7 
Oct 2005
Fribourg, Switzerlan
FC_{16} Posts 
1) I didn't said it was related to FLT
2) Thanks a lot for these specifications, really interresting 
20070516, 04:53  #8 
Jun 2003
3053_{8} Posts 
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? 
20070516, 09:31  #9 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 

20070516, 09:34  #10 
Nov 2003
2^{2}·5·373 Posts 

20070516, 09:51  #11 
Jun 2003
1,579 Posts 

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