From: elkies@math.harvard.edu (Noam Elkies)
Subject: x^4 + y^4 = z^4 + (3/4) t^4
Date: 24 Jun 00 16:43:54 GMT
Newsgroups: sci.math.numberthy
Summary: [missing]

The twisted Fermat quartic surface in the "Subject:" line
contains the rational curve

(x,y,z,t) = (192 a^7,  192 a^8 - 24 a^4 - 1,  192 a^8 + 24 a^4 - 1,  4 a)

As a consequence of this identity, not only does the surface have
infinitely many rational points (with >>N^(1/8) of height at most N,
rather than the expected N^epsilon), but there are infinitely many
solutions of  |x^4+y^4-z^4| << sqrt(z)  in nonzero integers  x,y,z
(where the usual naive heuristics suggest finitely many solutions
of  |x^4+y^4-z^4| < z^c  for each  c<1).

This latter application is how I found this identity, after seeing the
first few cases in a search for small |x^4+y^4-z^4| along the lines
explained here and in my ANTS-IV paper.  But I would not be surprised
to learn that the identity, or something equivalent to it, has been
known for a long time.  Does anybody here recognize it?

Thanks,
--Noam D. Elkies
==============================================================================

From: elkies@math.harvard.edu (Noam Elkies)
Subject: Correction Re:  x^4 + y^4 = z^4 + (3/4) t^4
Date: 25 Jun 00 03:17:52 GMT
Newsgroups: sci.math.numberthy

I wrote

>(x,y,z,t) = (192 a^7,  192 a^8 - 24 a^4 - 1,  192 a^8 + 24 a^4 - 1,  4 a)

>As a consequence of this identity, not only does the surface have
>infinitely many rational points (with >>N^(1/8) of height at most N,

Make that >>N^(1/4), actually; here $a$ need not be an integer, and
there are >>H^2 rationals of height at most H.

NDE
==============================================================================

From: Noam D. Elkies <elkies@maf'th.harvard.edu>
Subject: Re: An Obtuse Paradox?
Date: 4 Sep 2000 19:01:04 GMT
Newsgroups: rec.puzzles,sci.math

In article <39B07E16.6BB9@earthlink.net>,
Adam Stephanides  <adamsteph@earthlink.net> wrote:
>Noam D. Elkies wrote:

>> In article <39AFEFA5.7187@earthlink.net>,
>> Adam Stephanides  <adamsteph@earthlink.net> wrote:
>> >Whether what I originally showed was true of our set (that when x is
>> >irrational there are only a countable number of y such that (x,y) is in
>> >the set) implies that the set has measure zero is something I don't know.

>> This implication does not hold; in fact there are even non-measurable sets
>> that are graphs of functions, and thus contain a *single* (x,y) for each
>> choice of x!

>Suppose we assume, in addition to the property I stated, that the set is
>measurable.  Must its measure then be zero?  (My guess would be yes in
>this case.)

Good question, and you guess correctly.

Tile the plane with unit squares with sides parallel to the x and y axes,
and consider the intersection of our set with each one of these squares.
(I don't need to specify whether our squares our open or closed since
the boundary of each square is negligible.)  Each such intersection
is measurable, and the sum of their measures equals the measure of the
whole set.  [This is because we're dealing with a countable union that's
disjoint except on a negligible set.]  So, it's enough to show that
each intersection is negligible.

Let S, then, be a measurable subset of a unit square such that
for each x there are at most countably many y for which (x,y) is in S.
We claim that S is negligible.

If each x yields at most a single y (e.g. if S came from the graph of
a function) then this is easy because S then has infinitely many
disjoint translates (in the y-direction), all of which fit in the same
1*2 rectangle.  Thus if S had positive measure then the 1*2 rectangle
would have subsets of arbitrarily large measure, which is impossible.

In the general case of countably many y's, I don't see a slick way
of doing it, so I resort to the general theory of product measures
(see e.g. Dunford and Schwartz, _Linear Operators_ Vol.I, 3.11).
It is known that if S is a measurable subset of a unit square then,
for all x outside a negligible set, the set S_x of y such that (x,y)
is in S is a measurable subset of the real line; moreover, if f(x)
is the measure of S_x, then the measure of S is the Lebesgue integral
of f(x)dx.  In our case, S is known to be measurable, and each S_x
is at most countable, and therefore negligible.  Thus f(x)=0 for all x,
and the integral of f(x)dx is also zero.  So S is indeed negligible,
as claimed.

ObPuzzle: Suppose F is a family of negligible subsets of [0,1]
such that if S and S' are in F then either S contains S' or
S' contains S.  Must the union of all the sets in F be negligible?

--Noam D. Elkies <remove derivative of f from e-address to reply>
  Department of Mathematics, Harvard University
--Noam D. Elkies <change card(Z) to math in e-address to reply>
  Department of Mathematics, Harvard University