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 2007-05-11, 23:42 #1 Citrix     Jun 2003 64D16 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 2007-05-11 at 23:43
2007-05-12, 03:34   #2
wblipp

"William"
May 2003
Near Grandkid

2·1,187 Posts

Quote:
 Originally Posted by Citrix 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?

 2007-05-12, 15:05 #3 Citrix     Jun 2003 161310 Posts yes
 2007-05-15, 06:58 #4 nibble4bits     Nov 2005 18210 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 2007-05-15 at 06:59
2007-05-15, 16:56   #5
victor

Oct 2005
Fribourg, Switzerlan

22×32×7 Posts

Quote:
 Originally Posted by nibble4bits 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 $x, y, z \in [2; 500]$

2007-05-15, 21:31   #6
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

11101010110002 Posts

Quote:
 Originally Posted by victor 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

 2007-05-15, 22:11 #7 victor     Oct 2005 Fribourg, Switzerlan 22·32·7 Posts 1) I didn't said it was related to FLT 2) Thanks a lot for these specifications, really interresting
 2007-05-16, 04:53 #8 Citrix     Jun 2003 1,613 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?
2007-05-16, 09:31   #9
akruppa

"Nancy"
Aug 2002
Alexandria

9A316 Posts

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

Alex

2007-05-16, 09:34   #10
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

23·3·313 Posts

Quote:
 Originally Posted by Citrix 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.

2007-05-16, 09:51   #11
Citrix

Jun 2003

161310 Posts

Quote:
 Originally Posted by R.D. Silverman 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)

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