From: kevin2003@delphi.com (Kevin Brown)
Newsgroups: sci.math
Subject: The Sunday Telegraph's Puzzle
Date: 8 Jan 1995 00:25:03 GMT
It's been reported that the Sunday Telegraph of London had offered
a cash prize (L450) to the first person to send them a solution in
co-prime positive integers greater than 100 of the equation (A^3/B^3) +
(C^3/D^3) = 6. Many readers of this newsgroup will recognize this
as an old problem already "solved" in principle by Viete, Fermat,
Euler, etc. In fact, Diophantus could have solved it. Strangely,
Legendre once stated that this particular equation had no solutions,
but then Lame' pointed out (17/21)^3 + (37/21)^3 = 6.
I'm not sure the report was accurate (since I'm not a subscriber), but
the Telegraph must have started receiving solutions within minutes of
the report's appearance on the net. (On the chance that the offer
was genuine I sent them a solution about 30 minutes after the
problem appeared on 31 Dec, and presumably mine was not the first.)
The tangent-chord method, or some variation, yields all the solutions
I've seen posted so far. As Poincare conjectured and Mordell proved,
all the rational points on a curve such as x^3 + y^3 = 6 can be
generated by tangent and chord constructions applied to a finite
set of points. My question is, do we know the set of points that
generate all solutions in this particular case? All the solutions
I've seen so far were based on Lame's point (17/21,37/21).
What led Legendre to believe there were no solutions? Interestingly,
it can _almost_ be proved by descent that the equation can have no
solutions. Starting from scratch, we observe that (AD)^3 +
(BC)^3 = 6(BD)^3 implies that D^3 divides B^3 and that B^3 divides
D^3, so we have B=+-D and the equation can be written in the form
A^3 + B^3 = 6C^3 where A,B,C are integers.
Since A^3 + B^3 is even, whereas A and B cannot both be even,
they must both be odd, so we have co-prime integers u,v such that
A=u+v and B=u-v. Making these substitutions gives u(u^2 + 3v^2)
= 3C^3. Clearly u is divisible by 3. Also, since u and (u^2+3v^2)/3
are co-prime, they must both be cubes, so we have integers a,b such
that u = 27a^3 and u^2 + 3v^2 = 3b^3. From the theory of binary
quadratic forms it's known that all the solutions of the latter
equation are of the form
u = 9y(x^2 - y^2) v = x(x^2 - 9y^2) b = x^2 + 3y^2
where gcd(x,y)=1. Noting that u must be of the form 27a^3 we have
3a^3 = y(x+y)(x-y)
This shows that 3 must divide exactly one of the quantities y, (x+y),
or (x-y). In the first case it follows that (y/3), (x+y), and (x-y)
must all be cubes. Setting
(y/3) = q^3 (x+y) = r^3 (x-y) = s^3
we have (r)^3 + (-s)^3 = 6(q)^3, which is a solution of the original
equation in absolutely smaller integers. Therefore, if this were the
only possible case, we would be forced to conclude that no solution
is possible (by the principle of "infinite descent"). Legendre may
have been thinking along these lines when he stated the equation
was insoluable.
However, if y is not divisible by 3, then 3 divides either (x+y) or
(x-y), in which case we arrive at
(s)^3 + 2(q)^3 = 3(r)^3 or (r)^3 + 2(-q)^3 = 3(s)^3
respectively. At this point the "descent" argument no longer applies,
i.e., a solution to this equation does not necessarily imply an
absolutely smaller solution.
It follows that every solution of A^3 + B^3 = 6C^3 can be reduced
down through a sequence of successively smaller triples (q,r,s) that
satisfy the equation r^3 +s^3 = 6q^3 until reaching a triple that
satisfies instead the equation r^3 + 2s^3 = 3q^3. The smallest
solution (r,s,q) of the latter equation is (1,1,1), and the tangent
line through that point gives (-5,4,1). Every solution of the
Telegraph's problem that I've seen so far can be reduced to one of
these two irreducible cases. However, there are other solutions
that reduce to, for example, (655,-488,253), which is on the tangent
line through (-5,4,1).
Conversely, given a solution of either r^3 + s^3 = 6q^3 or r^3 +
2s^3 = 3q^3 we can generate the solution of A^3 + B^3 = 6C^3
"above it" by working backwards. For example, in the latter case
we would compute y=s^3 and x=3q^3 - s^3, and then the "higher"
solution of A^3+B^3=6C^3 is given by
A = 9y(x^2-y^2) + x(x^2-9y^2)
B = 9y(x^2-y^2) - x(x^2-9y^2)
Has anyone tried constructing a "family tree" of all the solutions
of A^3 + B^3 = 6C^3 generated from the irreducible solutions of the
related equation r^3 + 2s^3 = 3q^3 ?
==============================================================================
From: cet1@cus.cam.ac.uk (Chris Thompson)
Newsgroups: sci.math
Subject: Re: The Sunday Telegraph's Puzzle
Date: 8 Jan 1995 13:52:08 GMT
In article <9501071927591.kevin2003.DLITE@delphi.com>, kevin2003@delphi.com
(Kevin Brown) writes:
|> [...] As Poincare conjectured and Mordell proved,
|> all the rational points on a curve such as x^3 + y^3 = 6 can be
|> generated by tangent and chord constructions applied to a finite
|> set of points. My question is, do we know the set of points that
|> generate all solutions in this particular case? All the solutions
|> I've seen so far were based on Lame's point (17/21,37/21).
Yes, the group of rational points does have rank 1. One can grind this out
following the proof of Mordell's theorem as given in [1], say. Transform
X^3+Y^3=6Z^3 into Weierstrass form by x = 18Z/(X+Y), y = 27(X-Y)/(X+Y) to
get y^2 = x^3 - 3^5 which we want to show has a group G of rational points
generated by (7,10) alone. Now factorise over Q(d) where d=3^{1/3} and use
the isomorphism of G/2G into Q[d]/(Q[d])^2 given by mapping (x,y) -> x-3d^2
[1, pp 67+ff]. The argument goes like the example on [1, p.72]: x-3d^2 must
be the square of an ideal (3 ramifies completely), and as Q(d) has class
number 1 it must be a unit times the square of an element of Q(d). The
units are generated by -1 and d^2-2; using the real embedding allows only
x-3d^2 = l^2 or x-3d^2 = (d^-2)l^2, l in Q(d). Therefore G/2G has at most
two elements, and G has rank at most 1. In fact precisely 1, as 7-3d^2 =
(d^2-2)(d^2+1)^2 demonstrates the solubility of the second equation. A
height argument shows that (7,10) really is the generator of G, should
anyone need convincing.
Kevin gives a descent argument using the two curves
x^3 + y^3 + 6z^3 = 0 (A)
x^3 + 2y^3 + 3z^3 = 0 (B)
(flipping the sign of z for convenience), by which a point on A gives
either a point on A or one on B. This is essentially halving in G, with
the second case arising when the point is an odd multiple of the generator.
Now the argument on [1, p.86] shows how to lift a point on B back to A
(this is essentially the construction of the jacobian of B). For example,
the fact that (1,1,1) lies on B shows that (1+2w+3w^2,1+2w^2+3w,3), where
w^3=1, lies on A. Thus also P=(4+5w,5+4w,3) and its conjugate P' lie on A,
and so the third point of intersection of PP' with A is rational. Computing
its coordinates with Desboves' formulae [1,p.130] it turns out to be [suprise!]
(37,17,-21). This sort of descent + climbing back from B to A would give
another proof that the rank of G is 1.
[1] J.W.S. Cassels
Lectures on Elliptic Curves
London Math. Soc. student texts #24
Cambridge University Press (1991)
ISBN 0-521-41517-9 (hard) 0-521-42530-1 (soft)
Chris Thompson
Internet: cet1@phx.cam.ac.uk
JANET: cet1@uk.ac.cam.phx
==============================================================================
From: cet1@cus.cam.ac.uk (Chris Thompson)
Newsgroups: sci.math
Subject: Re: The Sunday Telegraph's Puzzle
Date: 8 Jan 1995 19:38:20 GMT
In article <3eoqm8$cdi@lyra.csx.cam.ac.uk>, I wrote:
[...]
|> Kevin gives a descent argument using the two curves
|>
|> x^3 + y^3 + 6z^3 = 0 (A)
|> x^3 + 2y^3 + 3z^3 = 0 (B)
|>
|> (flipping the sign of z for convenience),
but then messed up this sign reversal completely...
|> the fact that (1,1,1) lies on B shows that (1+2w+3w^2,1+2w^2+3w,3), where
should be: (1,1,-1) (1+2w-3w^2,1+2w^2-3w,3)
The rest is right (I think). Apologies.
Chris Thompson
Internet: cet1@phx.cam.ac.uk
JANET: cet1@uk.ac.cam.phx