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 February 20th, 2014, 11:06 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, 11:14 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 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, 11:26 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, 12:51 PM   #4
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Re: Beal's Conjecture Conjecture

Quote:
 Originally Posted by MrAwojobi 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.
Sorry to disappoint you [color=#0000BF]MrAwojobi[/color] , here's a counterexample:

$3000^5 \ + \ 10^{17} \= \ 700000^3$

I constructed it using as a seed,

$3^5 \ + \ 10^2 \= \ 7^3$

This just barely misses Beal criteria, exponent 2 is not allowed.

Last edited by skipjack; May 9th, 2014 at 04:41 AM.

February 20th, 2014, 01:08 PM   #5
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Re: Beal's Conjecture Conjecture

Quote:
 Originally Posted by MrAwojobi 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.
That is, if you don't have two exponents which are the same, you don't have an example. Right?

 February 20th, 2014, 01:21 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, 01:47 PM   #7
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Re: Beal's Conjecture Conjecture

I'm just trying to find out what you mean by
Quote:
 Originally Posted by MrAwojobi i.e. 2 of the 3 powers x,y or z will be the same
Would you clarify?

February 20th, 2014, 02:13 PM   #8
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Re: Beal's Conjecture Conjecture

Quote:
 Originally Posted by MrAwojobi agentredlum, you have not disappointed me because as you yourself acknowledge, your example simplifies to an equation that doesn't fulfil the Beal condition.
Then I don't understand what you are saying, every example of Beal can be 'simplified' to something that does not satisfy the Beal criteria, but it won't be the same number. Can you provide a counterexample that can't?

Take

$2^8 + 4^4= 2^9$

Where it 'seems' like the three exponents are all different but it is easy to manipulate this to

$2^8 + 2^8= 2^9$

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

$3000^5 + 10^{17}= 700000^3$

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 $\ \ 10^{15} \ \$ 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 04:43 AM.

 February 20th, 2014, 03:01 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, 03:36 PM #10 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 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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