# 05.24.07

## Beal’s Conjecture

Posted in math, programming at 9:47 pm by danvk

(I noticed today that Rice has finally taken down my owlnet page. Over the next few weeks, I’ll be giving some of the interesting pages from that site a new home here on danvk.org. First up is Beal’s Conjecture…)

I’ve put a link to last summer’s work on Beal’s Conjecture over on the right-hand side of the site. A quick overview:

Beal’s conjecture states that, if (x,y,z) are co-prime and m,n,r ≥ 3, then xm + yn ≠ zr. Sound familiar? It should. It’s a generalization of Fermat’s Last Theorem.

What makes Beal’s conjecture especially exciting is that Andrew Beal, a Texas billionaire, has put a \$100,000 prize on the proof or disproof of the problem. If it’s false (and most generalizations of FLT have been), then a computer search may have a chance of coming up with the counterexample. Peter Norvig did an initial hunt but came up empty. I extended his results, and also came up empty. Now that I have access to lots of machines, I’d like to extend the search a bit further.

The old article I wrote is still valid, though it contains a misstatement that’s made all the more embarrassing by being in ALL CAPS. I’ll be lazy and leave it as one of those pesky “exercises to the reader” to figure out how I goofed.

1. Aubrey de Grey said,

November 28, 2008 at 3:57 pm

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.

2. stan lesz said,

May 30, 2011 at 10:29 am

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.