From: Herman.te.Riele@cwi.nl
Newsgroups: sci.math.numberthy
Subject: Re: a question of Mordell
Date: 30 Nov 98 17:39:56 GMT

> Mordell asked whether there are infinitely many integral solutions of
> x^3 + y^3 + z^3 = 3.  Are there heuristic arguments that would lead one
> to expect this?  It strikes me as an unusual thing to conjecture, in
> view of the fact that as late as 1964, the only known solutions (known
> to Davenport, at least) were (1,1,1) and (4,4,-5) (and permutations).
>
> Jim Propp

See:

@article(HB2,
   author  = "D.R. Heath-Brown",
   title   = "The density of zeros of forms for which weak approximation fails",
   journal = MathComp,
   volume = "59",
   year    = "1992",
   pages = "613--623")

Related references:

@inproceedings(HB1,
   author  = "D.R. Heath-Brown",
   title   = "Searching for solutions of $x^3+y^3+z^3=k$",
   editor  = "D. Sinnou",
   booktitle = {S\'{e}m. {T}h\'{e}orie des {N}ombres, {P}aris 1989--1990},
   publisher = {Birkh\"{a}user},
   year = "1992",
   pages = "71--76")

@article(HBLR,
   author = "D.R. Heath-Brown and W.M. Lioen and H.J.J. te Riele",
   title = "On solving the {D}iophantine equation $x^3+y^3+z^3=k$
            on a vector computer",
   journal = MathComp,
   volume = "61",
   year = "1993",
   pages = "235--244")

See also pp. 324--327 of:

@article(Ref:SurCNT,
   author = "H.J.J.~te~Riele and J.~van~de~Lune",
   title = "Computational {N}umber {T}heory at {CWI} in 1970--1994",
   journal = "CWI Quarterly",
   year = "1994",
   volume = "7",
   number = "4",
   pages = "285--335")
==============================================================================

From: elkies@abel.math.harvard.edu (Noam Elkies)
Newsgroups: sci.math.numberthy
Subject: Re: a question of Mordell
Date: 30 Nov 98 17:39:57 GMT

Jim Propp writes:

> Mordell asked whether there are infinitely many integral solutions of
> x^3 + y^3 + z^3 = 3.  Are there heuristic arguments that would lead one
> to expect this?  It strikes me as an unusual thing to conjecture, in
> view of the fact that as late as 1964, the only known solutions (known
> to Davenport, at least) were (1,1,1) and (4,4,-5) (and permutations).

There are still no further solutions known.  The heuristics are that
the number of solutions with at most n digits should be asymptotic
to n.  The crudest explanation is that if x,y,z are "random" integers
the largest of whose absolute values lies in (N,2N] then we expect
x^3+y^3+z^3 to be a "random" number in [-AN^3,+BN^3], and we get CN^3
such random numbers, so the probability that one of them equals 3 is
the positive constant C/(A+B).  More refined heuristics take account of
the fact that x^3+y^3+z^3 is not uniformly distributed in that interval,
either in size or modulo small numbers (e.g. it can never be congruent
to 4 or 5 mod 9).  For instance it turns out that cubes are expected
to occur much more frequently as x^3+y^3+z^3, and this is borne out by
experiment.  [Twice-cubes also occur frequently, but this is entirely
accounted for by (6t^2)^3 + (6t^3-1)^3 - (6t^3+1)^3 = -2  and variations
on this identity.]  About two years ago I carried out the computation
of the expected distribution of solutions of Mordell's equation, and
found that, barring arithmetic errors, the expected number of
solutions of |x^3+y^3-z^3|=3 with 0<x<=y<z and z in [N,AN] should
be asymptotically log(A)*C(3) where C(3) is about .061.  That is,
one expects about n/7 solutions with at most n digits in z.  Thus
it is not surprising that no new ones have been found yet.  This
constant .061 is smaller by far (at least a factor of 2 if I remember
correctly) than the constant C(m) for x^3+y^3+z^3=m for any other
m between 1 and 100 (other than m congruent to 4 or 5 mod 9; when
m is a cube, the infinite product for C(m) diverges to infinity).
Thus x^3+y^3+z^3=3 should be the hardest such equation at least
with small m, and the tiny solutions (1,1,1) and (4,4,-5) seem
to be something of a fluke.

--Noam D. Elkies (elkies@math.harvard.edu)