From: ikastan@alumni.caltech.edu (Ilias Kastanas) Newsgroups: rec.puzzles,sci.math Subject: Re: cbrt comparison (Re: 163 magic) Date: 25 Oct 1997 10:02:23 GMT In article <62r689$cqq$1@news.fas.harvard.edu>, Noam Elkies wrote: >In article <3450EB37.64E0@mail.idt.net.REMOVE>, >Tenie Remmel wrote: >>Noam Elkies wrote: >>> ObPuzzle: which of cbrt(186919) and cbrt(31226) + cbrt(16948) Let us set a = 186919, b = 31226, c = 16948. >>> is larger (cbrt being the cube-root function)? >> >>easy if you have a 10 digit calculator with 5 hidden digits. >> >>cbrt(186919) = 57.1765328287656+ >>cbrt(31226) + cbrt(16948) = 57.1765328287650+ > >Assuming the numbers are in fact distinct, the difference must >of course become apparent with enough hidden digits, but 10+5 >is not enough: in the above, both numbers are wrong in the last >place, the correct digit being 9+ for cbrt(186919) as well as >cbrt(31226) + cbrt(16948). Suppose (at)^1/3 + b^1/3 + c^1/3 = 0; then at + b + c = 3(tabc)^1/3, so t is a root of P(x) = (ax + b + c)^3 - 27abcx. We can compare t to -1 using integer arithmetic: P(-1) = 27abc - (a - b - c)^3, and this is bound to be small by the nature of the puzzle. Since it is -1 mod 2, 3 and 5 a reasonable guess is P(-1) = -1; which is easy to verify when written as 27abc = (a - b - c -1)((a - b - c)^2 + a - b - c +1). E.g. a - b - c -1 = = 138744 = 8*9*1927 while a = 97*1927. By this simple approach -1 < t < 0, and a^1/3 > b^1/3 + c^1/3. Ilias ============================================================================== From: elkies@ramanujan.harvard.edu (Noam Elkies) Newsgroups: rec.puzzles,sci.math Subject: *Spoilers* Re: cbrt comparison [not 163] Date: 27 Oct 1997 00:17:34 GMT In article <62sg3f$l5n@gap.cco.caltech.edu>, Ilias Kastanas wrote: [...] >>>Noam Elkies wrote: >>>> ObPuzzle: which of cbrt(186919) and cbrt(31226) + cbrt(16948) >>>> [is greater?] > Let us set a = 186919, b = 31226, c = 16948. > Suppose (at)^1/3 + b^1/3 + c^1/3 = 0; then at + b + c = 3(tabc)^1/3, >so t is a root of P(x) = (ax + b + c)^3 - 27abcx. We can compare t to >-1 using integer arithmetic: P(-1) = 27abc - (a - b - c)^3, and this is bound >to be small by the nature of the puzzle. Since it is -1 mod 2, 3 and 5 a >reasonable guess is P(-1) = -1; which is easy to verify when written as >27abc = (a - b - c -1)((a - b - c)^2 + a - b - c +1). E.g. a - b - c -1 = >= 138744 = 8*9*1927 while a = 97*1927. > > By this simple approach -1 < t < 0, and a^1/3 > b^1/3 + c^1/3. And snowe@rain.org (Nick Halloway) and Kurt Foster (kfoster@rmi.net) also obtained the (a-b-c)^3-27abc condition, with Kurt F. finding the nice interpretation in terms of the determinant X^3+Y^3+Z^3-3XYZ of a 3x3 circulant matrix. There's little I can add to these solutions except for the fact that (a-b-c)^3-27abc must a priori be congruent to 0, 1, or -1 mod 9. Ilias K.'s factorization trick for verifying "P(-1)=-1" also goes part of the way towards answering the question asked by Nick H. (and implicitly also by Kurt F. in e-mail): >So how are these things dreamed up? 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... --Noam D. Elkies (elkies@math.harvard:edu) Dept. of Mathematics, Harvard University ==============================================================================