Comments on: Beal’s Conjecture http://www.danvk.org/wp/2007-05-24/beals-conjecture/ Keepin' static like wool fabric since 2006 Sun, 05 Feb 2012 04:54:46 +0000 hourly 1 http://wordpress.org/?v=3.3.1 By: stan lesz http://www.danvk.org/wp/2007-05-24/beals-conjecture/comment-page-1/#comment-83437 stan lesz Mon, 30 May 2011 17:29:13 +0000 http://www.danvk.org/wp/?p=158#comment-83437 Beal's m,n,r=>3 3^8-18^3=3^6 3^14-162^3=3^12 104^3-91^3=13^5 1352^3-1183^3=13^8 ??? Flt is solvable in Gaussian numbers. (a+√b)^3-(-a+√b)3=c^3 (3+√93)^3-(-3+√93)^3=12^3 Summation of [(n+1)^3-n^3]=(c+1)^3 n=0 to c to upper limit. Beal’s m,n,r=>3
3^8-18^3=3^6
3^14-162^3=3^12
104^3-91^3=13^5
1352^3-1183^3=13^8
??? Flt is solvable in Gaussian numbers.
(a+√b)^3-(-a+√b)3=c^3
(3+√93)^3-(-3+√93)^3=12^3

Summation of [(n+1)^3-n^3]=(c+1)^3
n=0 to c to upper limit.

]]> By: Aubrey de Grey http://www.danvk.org/wp/2007-05-24/beals-conjecture/comment-page-1/#comment-15004 Aubrey de Grey Fri, 28 Nov 2008 22:57:53 +0000 http://www.danvk.org/wp/?p=158#comment-15004 Dan - have you considered adapting your program to seek additional solutions to the weaker problem where one of the exponents can be 2 but the sun of their reciprocals must be less than 1? The known solutions are intriguingly distributed: apart from the semi-trivial solution 1+8=9, there are just four solutions to this in which z^r is under 15,000 and then five more where it is between 10^11 and 10^15. No others are known, but only very weak partial non-existence results have been proved (the strongest being that there are only finitely many solutions for any particular mnr triplet), and I'm not sure that any search has gone very high. Regarding the Beal conjecture itself, a search in which the exponents are all kept really low (<20) seems more interesting - the density of such powers is so much higher than that of high powers that you'd be doing most of your existing search, and you could take the bases up to very large numbers in compensation. Dan – have you considered adapting your program to seek additional solutions to the weaker problem where one of the exponents can be 2 but the sun of their reciprocals must be less than 1? The known solutions are intriguingly distributed: apart from the semi-trivial solution 1+8=9, there are just four solutions to this in which z^r is under 15,000 and then five more where it is between 10^11 and 10^15. No others are known, but only very weak partial non-existence results have been proved (the strongest being that there are only finitely many solutions for any particular mnr triplet), and I’m not sure that any search has gone very high.

Regarding the Beal conjecture itself, a search in which the exponents are all kept really low (<20) seems more interesting – the density of such powers is so much higher than that of high powers that you’d be doing most of your existing search, and you could take the bases up to very large numbers in compensation.

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