From - Tue Nov 4 10:37:15 1997 From: elkies@ramanujan.harvard.edu (Noam Elkies) Newsgroups: rec.puzzles,sci.math Subject: Re: cube root question was: Re: 163 magic ... Date: 4 Nov 1997 04:32:33 GMT Organization: Harvard Math Department Lines: 76 Message-ID: <63m8h1$4lf$1@news.fas.harvard.edu> References: <19971022171801.NAA29091@ladder02.news.aol.com> <3450237B.1B3D@mail.interport.net> <62p2q7$q14$1@news.fas.harvard.edu> In article , Christopher J. Henrich wrote: >In article <62p2q7$q14$1@news.fas.harvard.edu>, >elkies@ramanujan.harvard.edu (Noam Elkies) wrote: >> ObPuzzle: which of cbrt(186919) and cbrt(31226) + cbrt(16948) >> is larger (cbrt being the cube-root function)? >Well, I can answer this, but I can also see that something deep was going on >behind the cooking-up of the question. I don't know how "deep" I'd claim this is, but certainly those numbers were not just made up randomly... I explained how I got them in a recent post, from which I quote at the end. >If (x,y,z) are homogeneous coordinates on the real projective plane, >then the equation x^1/3 + y^1/3 - z^1/3 = 0 defines a curve. After some >manipulation you can rationalize that equation to this: >Q(x,y,z) = (x + y - z)^3 + 27xyz = 0. >the curve Q(x,y,z)=0 is not elliptic, but rational. It has a singularity >at (-1, -1, 1). Right -- but then you knew already that it had a rational parametrization, to wit (x,y,z)=(1,t^3,(1+t)^3). >The theory presumably has to do with values of cubic forms. I hazard >a guess that Q(x,y,z) is the norm of some algebraic number, and something >about some ideal hinges on Q having a small value. It is true that Q(x,y,z) is the norm of x^(1/3)+y^(1/3)-z^(1/3), which is one explanation of why we want to make Q(x,y,z) small but didn't help me actually find x,y,z which solve Q(x,y,z)=1. --Noam D. Elkies (elkies@math.harvard.edu) Dept. of Mathematics, Harvard University ===================================================================== "I calculated this solution of |(a-b-c)^3-27abc|=1 in late 1981, "in response to the challenge cbrt(60) ? 2 + cbrt(7), trying "to find out how close cbrt(a) can get to cbrt(b)+cbrt(c) "without equaling it. I found an infinite series of triples "making the difference as small as possible relative to the "size of a,b,c, of which (186919;16948,31226) was the first. "This story is told in Crux Mathematicorum 1982. " "I reduced the problem to a family of inhomogeneous quadratic "equations in two integer variables, and used the continued "fractions approach to (Fermat-)Pell equations to find a solution "to one of those quadratic equations to find the first solution, "from which the existence of infinitely many solutions followed. " "My original reduction was rather baroque, but one simpler way to see "it is to write the equation symmetrically as (a+b+c)^3 - 1 = 27abc, "and let r = (a+b+c-1)/a. For each rational value of r we "get the quadratic equation (a+b+c)^2+a+b+c+1 = 27bc/r, which "[since (r-1)a=b+c-1] is in effect a quadratic in two variables. " "Only much later did I run an exhaustive computer search for "solutions, finding that the minimal triple is (313;14,84). "But I did not pose the question of comparing cbrt(313) with "cbrt(14)+cbrt(84) because there the difference of about 6.76e-08 "is barely small enough to detect with a 10-digit calculator, "whereas (186919;16948,31226) yields 2.346e-15, safely beyond "calculator range. There are a couple of other smaller solutions: "(46879; 4996, 6818) and (86830; 4881, 20388). I used my original "186919 solution to pose the puzzle for sentimental reasons. " "This all seems in retrospect a remarkable prefiguration of the "story of a^4+b^4+c^4=d^4, whose earliest (but as it turned out "later not smallest) nontrivial solution I found about six years "afterwards using a parametrization by curves of genus 1. To be "sure the |(a-b-c)^3-27abc|=1 problem did not have the distinguished "pedigree of the exponent-4 case of Euler's 1769 conjecture...