superprismatic Posted November 19, 2011 Report Share Posted November 19, 2011 BEAL'S CONJECTURE: If Ax + By = Cz , where A, B, C, x, y and z are positive integers and x, y and z are all greater than 2, then A, B and C must have a common prime factor. EXAMPLE: 36 + 183 = 38 COOL THING: As far as I can tell, a prize of $100,000 is still unclaimed for someone who can produce either a proof or a counterexample of this conjecture. Does anyone have any ideas? I'd love to see someone from this forum win the prize! Quote Link to comment Share on other sites More sharing options...
0 Faizaan Posted November 19, 2011 Report Share Posted November 19, 2011 COOL THING: As far as I can tell, a prize of $100,000 is still unclaimed for someone who can produce either a proof or a counterexample of this conjecture. $100,000. Still not enough motivation to get me to sit down and try to figure this out. My laziness know no bounds. Anywho, I had not heard of this conjecture until now. This is intriguing! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 19, 2011 Report Share Posted November 19, 2011 I'm having a really hard time showing any case where the values aren't all divisible by 2 or 3. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 19, 2011 Report Share Posted November 19, 2011 (edited) Is x=y=z? edit: never mind... i didnt see the example. Edited November 19, 2011 by Assassinator Quote Link to comment Share on other sites More sharing options...
0 EventHorizon Posted November 19, 2011 Report Share Posted November 19, 2011 Well, if any two of {A,B,C} have the same prime factor, then the other must (trivial to prove). So we are looking for a case where all 3 are relatively prime. I'm thinking my approach will involve lots of modular arithmetic, but nothing of note to report right now. And given there's a prize on this one, I'm thinking it's been tried before by people more familiar with the necessary maths to solve it (I know next to nothing about rings/fields/etc). Still interesting though, so I'll play with it for a while. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 19, 2011 Report Share Posted November 19, 2011 One possible answer A = 3 B = 3 C = 6 All have common prime factor of 3 X = 2 Y = 3 Z = 2 X, Y and Z are positive integers greater than or equal to 2 32 + 33 = 62 = 36 Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted November 19, 2011 Author Report Share Posted November 19, 2011 One possible answer A = 3 B = 3 C = 6 All have common prime factor of 3 X = 2 Y = 3 Z = 2 X, Y and Z are positive integers greater than or equal to 2 32 + 33 = 62 = 36 x,y, and z must all be greater than 2. Quote Link to comment Share on other sites More sharing options...
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superprismatic
BEAL'S CONJECTURE: If Ax + By = Cz , where A, B, C, x, y and z are positive integers
and x, y and z are all greater than 2, then A, B and C must have a common prime factor.
EXAMPLE: 36 + 183 = 38
COOL THING: As far as I can tell, a prize of $100,000 is still unclaimed for someone
who can produce either a proof or a counterexample of this conjecture.
Does anyone have any ideas? I'd love to see someone from this forum win the prize!
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