Modified fermat's last theorem

Citrix · 2007-05-11, 23:42

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?

wblipp · 2007-05-12, 03:34

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?

Citrix · 2007-05-12, 15:05

yes

nibble4bits · 2007-05-15, 06:58

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?

victor · 2007-05-15, 16:56

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 $x, y, z \in [2; 500]$

R.D. Silverman · 2007-05-15, 21:31

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
leads to the impossibility.

victor · 2007-05-15, 22:11

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

2) Thanks a lot for these specifications, really interresting

Citrix · 2007-05-16, 04:53

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?

akruppa · 2007-05-16, 09:31

Quote:

Originally Posted by victor

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

No, but Citrix originally did. See thread title.

Alex

R.D. Silverman · 2007-05-16, 09:34

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.

Citrix · 2007-05-16, 09:51

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)

2007-05-11, 23:42	#1
Citrix Jun 2003 7·233 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^xy^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*

2007-05-15, 06:58	#4
nibble4bits Nov 2005 2×7×13 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*

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2007-05-15, 22:11	#7
victor Oct 2005 Fribourg, CH 252₁₀ 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 7·233 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^la and y=p^mb and z=p^nc we have lx+my=nz How does one proceed from here?