Computing holowness by dr Perelman and Andrew Beal’s conjecture

Andrew Beal, http://www.pokerlistings.com/poker-player_andy-beal, conjectured http://www.bealconjecture.com/. The conjecture became interesting among many, including Peter Norvig (Director of Research, Google) http://norvig.com/beal.html. No other colaborative mathematical research, online, available, open. Yes, prizes do that.

Take the proposition that A^x+B^y=c^z is possible only when A,B,C have a common divisor. If we want to try a counterexample, we’ll try out A=kA_1, B=kA_2, C=A_3, where A_1,A_2,A_3 don’t divide k. WLOG y>x, LHS=k^x(A_1^x+A_2^{x+m}+k^m), where m=y-x. Contradiction. The same for A=kA_1, B=A_2, C=kA_3, as RHS=C^z-A^x=(kA_3)^{x+m}-(kA_1)^x=k^x(k^mA_3{x+m}-(kA_1)x=k^x(k^mA_3^{x+m}-A_1^x)\rightarrow k^x|RHS\rightarrow k|RHS\rightarrow contradiction.

Then, we have another situation, when A,B,C have no common divisor. It’s trivial to show that two of the numbers must be odd and one odd. It’s trivial to show that in order to prove the initial theorem we need to disprove the existence of triples such that A^x+B^y=C^z for A,B odd and C even as well as triples such that B^y=C^z-A^x, for odd A,C and even B.

The initial observation is the one connected to Pithagorean triples and infinite descent, ie. trying to order the potential solutions and finding out that the one solution would imply the existence of yet another one, which, though, would be leading to contradiction.

This is colaborative mathematics. If you find leaks (yes, Vinay Deolalikar and Andrew Wiles also did theirs), erros etc. – please, feel free to discuss. I want to change my opinion as long it’s not a perfect one.

 

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About misha

Imagine a story that one can't believe. Hi. Life changes here. Small things only.
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One Response to Computing holowness by dr Perelman and Andrew Beal’s conjecture

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