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: 0719383437 (add country prefix?)
Telephone: 18003678313 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 58digit 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 email 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^3z^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 email 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^3z^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 allpositive coordinates as above. But 6P also has
allpositive coordinates:
x = 1498088000358117387964077872464225368637808093957571271237
y = 1659187585671832817045260251600163696204266708036135112763
z = 1097408669115641639274297227729214734500292503382977739220
and this solution is in fact the smallest allpositive 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 email.)
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 allpositive
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