From: 70372.1170@COMPUSERVE.COM (Harvey Dubner) Newsgroups: sci.math.numberthy Subject: Puzzle announcement Date: 1 Jan 95 05:00:11 GMT Here is a chance for a number theorist to make some money. The Sunday Telegraph of London has an annual New Year Quiz covering many different subjects. Question 9, which is based on Question 8, is a number theory problem and has a special prize of 450 Pounds for the first person to submit the correct answer by January 11. Send answers to: Adrian Berry's Scope Quiz The Sunday Telegraph 1 Canada Square London E14 5AP Fax: 071-938-3437 (add country prefix?) Telephone: 1-800-367-8313 x6476 or x6477 Email: scope@telegraph.co.uk Here are copies of the questions: 8. Solve the equation, A^3/B^3 + C^3/D^3 = 6 where A, B, C, D are all positive whole numbers below 100. 9. A special question with a L450 prize. Either give a second solution to the above equation where the four variables are all whole numbers above 100 (A,B and C,D relatively prime), or demonstrate that no such second solution can exists. ============================================================================== Newsgroups: sci.math From: pmontgom@cwi.nl (Peter L. Montgomery) Subject: Re: number theory puzzle Date: Tue, 3 Jan 1995 05:21:50 GMT In article Laura Helen writes: >Something I got from the number theory mailing list, from Harvey Dubner: >>8. Solve the equation, A^3/B^3 + C^3/D^3 = 6 where A, B, C, D are all >>positive whole numbers below 100. >>9. A special question with a L450 prize. Either give a second solution >>to the above equation where the four variables are all whole numbers >>above 100 (A,B and C,D relatively prime), or demonstrate that no such >>second solution can exists. >Well, the first equation has the solution A=17, C=37, B=D=21. And no others >I think in naturals below 1000. > >I am interested in your thoughts on solving this equation. A computer search may barely find 6 = (2237723/960540)^3 - (1805723/960540)^3 , but this does not qualify because the integers are not all positive. I found a solution with 58-digit positive integers, but will let others work on it before I announce the solution. -- Peter L. Montgomery pmontgom@cwi.nl Mathematically gifted, unemployed, U.S. citizen. Interested in computer architecture, program optimization, computer arithmetic, cryptography, compilers, computational mathematics. 17 years industrial experience. ============================================================================== From: LAURAHELEN@news.delphi.com (LAURAHELEN@DELPHI.COM) Newsgroups: sci.math Subject: Re: number theory puzzle Date: 3 Jan 1995 12:44:10 -0500 LAURAHELEN@news.delphi.com (LAURAHELEN@DELPHI.COM) writes: >>you can simplify it to solving X^3 + Y^3 = 6 Z^3, X,Y,Z relatively prime >>naturals. >It was pointed out to me by e-mail that in Ireland and Rosen's >"Classical Introduction to Modern Number Theory" there is a theorem >that if x^3 + y^3 = az^3 has integer solutions it has infinitely many: >x1 = x(x^3+2y^3) >y1 = -y(2x^3 + y^3) >z1 = z(x^3-z^3) >and gcd(x1,y1,z1) = 1 or 3, gives another integer solution. You have to apply this 3 times to get solutions x3 = 7.92220574 exp 101 y3 = 6.779598061 exp 100 z3 = 4.360668424 exp 101 which have gcd 1. Thanks to the person who pointed out you have to do this 3 times not two. Laura ============================================================================== From: cet1@cus.cam.ac.uk (Chris Thompson) Newsgroups: sci.math Subject: Re: number theory puzzle Date: 4 Jan 1995 15:21:32 GMT In article <3ec2da\$cc5@news2.delphi.com>, Laura (LAURAHELEN@DELPHI.COM) writes: |> >>you can simplify it to solving X^3 + Y^3 = 6 Z^3, X,Y,Z relatively prime |> >>naturals. |> |> >It was pointed out to me by e-mail that in Ireland and Rosen's |> >"Classical Introduction to Modern Number Theory" there is a theorem |> >that if x^3 + y^3 = az^3 has integer solutions it has infinitely many: |> |> >x1 = x(x^3+2y^3) |> >y1 = -y(2x^3 + y^3) |> >z1 = z(x^3-z^3) |> |> >and gcd(x1,y1,z1) = 1 or 3, gives another integer solution. |> |> You have to apply this 3 times to get solutions |> |> x3 = 7.92220574 exp 101 |> y3 = 6.779598061 exp 100 |> z3 = 4.360668424 exp 101 |> |> which have gcd 1. Thanks to the person who pointed out you have to do this |> 3 times not two. This "tangent process" is essentially doubling under the Abelian group composition of points on the elliptic curve. From P=(37,17,21) you have found 8P which has all-positive coordinates as above. But 6P also has all-positive coordinates: x = 1498088000358117387964077872464225368637808093957571271237 y = 1659187585671832817045260251600163696204266708036135112763 z = 1097408669115641639274297227729214734500292503382977739220 and this solution is in fact the smallest all-positive one after P itself. For this you need to use the "chord process" as well as the "tangent process": see Cassel's "Lectures on Elliptic Curves" for a good introduction to the subject. (The attributions on p.24 --- the "tangent process" to Newton and the "chord process" to Diophantos --- are incorrectly transposed, as Prof. Cassels himself pointed out to me recently in e-mail.) This thread is reminiscent of the posting I made in sci.math a couple of months ago about H.E.Dudeney's treatment of X^3 + Y^3 = 9 Z^3 in "The Canterbury Puzzles". In this case also the first all-positive solutions after the generator P=(2,1,1) are 6P (claimed as new by Dudneney) and 8P (attributed to Fermat). This is a consequence of the fact the elliptic logarithms (or equivalently, the ratio x/y) for the generators in the two cases are close to each other. Incidentally, in the solutions section of Dudeney's book he writes | Some years ago I published a solution for the case of | | 6 = (17/21)^3 + (37/21)^3 | | of which Legendre gave at some length a "proof" of impossibility; | but I have since found that Lucas anticipated me in a communication | to Sylvester. Can anyone provide more details of this mistaken "proof" of Legendre? Chris Thompson Internet: cet1@phx.cam.ac.uk JANET: cet1@uk.ac.cam.phx