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...