February 20th, 2014, 10:06 AM  #1 
Senior Member Joined: Aug 2010 Posts: 158 Thanks: 4  Beal's Conjecture Conjecture
I conjecture that all examples of the Beal Conjecture equation A^x + B^x = C^z where A,B.C.x,y,z are positive integers and x>2, y>2, z>2 can be transformed to a Beal Conjecture equation of 5 variables and not 6 i.e. 2 of the 3 powers x,y or z will be the same. Can anyone supply the proof or a counterexample ?

February 20th, 2014, 10:14 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Beal's Conjecture Conjecture
If I understand you correctly, you're saying that if x, y, and z are distinct integers greater than 2 then there are no positive integers A, B, and C such that A^x + B^y = C^z. Is that right?

February 20th, 2014, 10:26 AM  #3 
Senior Member Joined: Aug 2010 Posts: 158 Thanks: 4  Re: Beal's Conjecture Conjecture
What you are saying is not what I am saying. I am simply conjecturing that all the Beal equation examples can be rewritten such that 2 powers out of 3 will be the same.

February 20th, 2014, 11:51 AM  #4  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: Beal's Conjecture Conjecture Quote:
I constructed it using as a seed, This just barely misses Beal criteria, exponent 2 is not allowed. Last edited by skipjack; May 9th, 2014 at 03:41 AM.  
February 20th, 2014, 12:08 PM  #5  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Beal's Conjecture Conjecture Quote:
 
February 20th, 2014, 12:21 PM  #6 
Senior Member Joined: Aug 2010 Posts: 158 Thanks: 4  Re: Beal's Conjecture Conjecture
agentredium, you have not disappointed me because as you yourself acknowledge, your example simplifies to an equation that doesn't fulfil the Beal condition. CRGreathouse what I am conjecturing is that every example of the Beal equation will always be manipulated to give another Beal equation via multiplication, division or rewriting a P^Q = R^S such that eventually 2 of the powers will be the same.

February 20th, 2014, 12:47 PM  #7  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Beal's Conjecture Conjecture
I'm just trying to find out what you mean by Quote:
 
February 20th, 2014, 01:13 PM  #8  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: Beal's Conjecture Conjecture Quote:
Take Where it 'seems' like the three exponents are all different but it is easy to manipulate this to These are the SAME number (2^9) and two out of three exponents are equal. Note that both representations above of 2^9 satisfy the Beal criteria but you cannot manipulate and get the same number (700000^3) with two out of three exponents equal. My counterexample satisfies the Beal criteria for the bases and exponents and there is a common factor of so it is NOT a counterexample to Beal. It is a counterexample to your claim however, IFF I understand your claim correctly. If you do not require that the number be kept the same after the transformation, then your conjecture is true. Last edited by skipjack; May 9th, 2014 at 03:43 AM.  
February 20th, 2014, 02:01 PM  #9 
Senior Member Joined: Aug 2010 Posts: 158 Thanks: 4  Re: Beal's Conjecture Conjecture
agentredium your last statement is correct. It seems you now agree with my conjecture. Can you prove it? CRGreathouse I think agentredium has provided a good explanation of what I am on about.

February 20th, 2014, 02:36 PM  #10 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Beal's Conjecture Conjecture
agent, if you do understand this, would you translate please?


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