Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10.2 (MU) 9/23/84; site aaec.OZ
Path: utzoo!watmath!clyde!burl!ulysses!gamma!epsilon!zeta!sabre!petrus!bellcore!decvax!genrad!panda!talcott!harvard!seismo!munnari!basser!aaec!rpb
From: rpb@aaec.OZ (Bob Backstrom)
Newsgroups: net.math
Subject: Re: Riemann Hypothesis (Actually Falting & Mordell Conjecture)
Message-ID: <484@aaec.OZ>
Date: Wed, 4-Sep-85 23:15:17 EDT
Article-I.D.: aaec.484
Posted: Wed Sep  4 23:15:17 1985
Date-Received: Sat, 7-Sep-85 04:28:34 EDT
References: <3129@nsc.UUCP> <616@petsd.UUCP>, <105@milford.UUCP> <1015@CS-Arthur>
Organization: Australian Atomic Energy Commission
Lines: 60

> Gerd Faltings' proof of Mordell's conjecture has apparently been
> accepted by the mathematical community, though I haven't seen it.  As a
> particular application of Mordell's conjecture, there can be at most
> finitely many solutions to the equation  x^n + y^n = z^n  in relatively
> prime integers x, y, and z, for a given n > 3.  Fermat's "theorem"
> claims that there are *no* solutions.

  
[ Another posting to the net since mail to jwt@CS-Arthur.ARPA was returned
with Host unknown message. ]


It's amazing how close Fermat's Theorem actually comes to being false.

Consider the following sums of cubes:
  
     3       3        3
    6 +     8  =     9  - 1

     3       3        3
   71 +   138  =   144  - 1
  
     3       3        3
  135 +   138  =   172  - 1
  
     3       3        3
17328 + 27630  = 29737  - 1
 
etc. etc.
  
as well as sums with + 1 on the r.h.s.:

     3       3        3
  577 +  2304  =  2316  + 1
 
     3       3        3
13294 + 19386  = 21279  + 1
  
  
etc.


There are infinitely many such occurrences as can be shown by the
following three algebraic identities:
         
              3     3      4 3      4      3
(1)        (9n  + 1)  + (9n )  = (9n  + 3n)  + 1

              3     3      4      3      4 3
(2)        (9n  - 1)  + (9n  - 3n)  = (9n )  - 1
 
              2 3      3     3      3     3
(3)        (6n )  + (6n  - 1)  = (6n  + 1)  - 2
  
Not all of the above examples can be generated from formulas (1) 
and (2), so the problem is to find ALL the "near" Fermat solutions.

  
From Bob Backstrom, Australian Atomic Energy Commission,
Sydney, New South Wales, Australia.