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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

445.0. "Fermat's last theorem" by NMGV01::ASKSIMON (Simon Clinch) Sat Mar 01 1986 16:54

  Referring to 432.0-.10,  a separate subject that I think needs
  some air-time is the non-iterative approach to Fermat's last
  theorem.
  
  So far,  noone has shown that it's unsoluble (or have they?).
  But noone has solved it either.
  
  The most obvious first step I think is to put:

         n   n   n
  	y = z - x 	(n>2)

  then...
  
  	 n	    n-1	   n-2 	          n-1
  =>	y = (z-x) (z    + z   x  + ... + x   )
  
  Now the number theory starts,  which is what Fermat was
  good at and therefore seems like the best way to go,  it
  being that he reckoned it was soluble.
  						       n
  The first clear requirement is that z-x must divide y.
  When I tackled the problem a few years ago,  I managed to
  prove that z-x is not divisible by any power of y,  and does
  not equal 1.

  So what next?  Well we can put,
  
  	y = y y ... y
  	     1 2     m

  where all these factors are prime by one of Fermat's *proved*
  theorems.  ("Unique factorization into primes" - Easy to prove).

  So clearly z-x must have a missing factor in this list,  which
  w.l.g. we shall call y .  Now we have the requirement that:
  		        1

	 n     n-1   n-2	   n-1
	y  | (z   + z   x + ... + x   )
	 1

  So this is all that remains to prove (or disprove) Fermat's last
  theorem along the steps I have taken. I.e.  does assuming this last
  requirement lead to a contradiction?
  
  If anyone can find a way of dealing with this,  or can see any
  promising lines to take,  I'd be grateful!  
T.RTitleUserPersonal
Name
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445.1First look at what CAN be provedMETOO::YARBROUGHMon Mar 03 1986 10:0112
    The problem with this approach is that the size of N for which it
    can be proved that no solutions can exist is prohibitively large.
    From Beiler's "Recreations in the Theory of Numbers", p. 288:
    
    	"...in 1941 it was shown that if x^p+y^p+z^p = 0 and no one
    of x, y, or z is a multiple of p, the p must be not less than
    253747889."
    
    Since the multiplicity condition in this statement is fairly easy
    to prove, we are already at the limits of the storage capacity of
    our largest computers just to store the coefficients of the polynomial
    which is to be factored.
445.2Still bigger lower limits...26608::YARBROUGHWhy is computing so labor intensive?Thu May 05 1988 15:166
Andrew Granville and Michael Monagan (Univ. of Saskatchewan) recently
extended the exponent in the previous note to 714,591,416,091,389. They 
used MAPLE extensively in the proof, including factoring some integers of 
153 digits, and evaluating the determinants of matrices of order 43 whose 
elements are univariate polynomials of order up to 50. Not the sort of 
thing you do on the back of an envelope!
445.3I'll take the setupPOOL::HALLYBYou have the right to remain silent.Thu May 05 1988 15:215
> 153 digits, and evaluating the determinants of matrices of order 43 whose 
> elements are univariate polynomials of order up to 50. Not the sort of 
> thing you do on the back of an envelope!

    ... or the margin of a book
445.4I got a good chuckle out of this, tooVMSDEV::HALLYBThe Smart Money was on GoliathFri Nov 03 1989 09:0021
From: [email protected] (Andrew P. Mullhaupt)
Newsgroups: sci.math
Subject: Re: number theory proofs?
Organization: Morgan Stanley & Co. NY, NY
 
In article <[email protected]>, [email protected] (H. de Bruijn) writes:
> When I was young, NOBODY could present a proof of Fermat's Last Theorem:
>  |
>  | There exist NO non-zero integers  x, y, z  such that:
>  |
>  |   x^3 + y^3 = z^3
>  |
> 
> Or am I not well informed anymore ?!?
Logically speaking, this depends on when you were young. You had to either
have been young a very long time ago, (Ahh, I'm not exactly sure but I think
certainly you would be a contemporary of Euler - in which case - you have
my sincere congratulations!), or the latter.
 
Later,
Andrew Mullhaupt
445.5for all practical purposes ;-)CIVAGE::LYNNLynn Yarbrough @WNP DTN 427-5663Mon Jan 13 1992 15:327
>Andrew Granville and Michael Monagan (Univ. of Saskatchewan) recently
>extended the exponent in the previous note to 714,591,416,091,389. 

... and this figure has recently been extended to 7.56E17, by Coppersmith
["Fermat's Last Theorem (Case 1) and the Wieferich Criterion", Mathematics 
of Computation Vol 54 pp.743-750 (1989)]. Reported in the MAPLE Technical 
Newsletter, Fall 1991.
445.7TRACE::GILBERTOwnership ObligatesWed May 06 1992 12:2810
Wow!  Math moves by leaps and bounds.  The 1991 result by Buhler, Crandall,
and Sompolski was (only) that FLT is true for n <= 10^6.  Now it's 7.56�10^17.

Using Inkeri's 1953 result: if (x,y,z) is a primitive counterexample to FLT
with exponent p (p prime) then x > [(2p�+p)/log10(3p)]^p.  When combined with
the result n <= 7.56�10^17, this means x must have at least 3.982�10^19 digits!


Also, in 1983 Faltings proved that for every n >= 3, fermat's equation has only
finitely many primitive solutions (unlike the generators of Pythagorean triples).
445.8GUESS::DERAMODan D&#039;Eramo, zfc::deramoWed May 06 1992 12:447
        re .7,
        
        From .1 and "(Case 1)" in .5, it appears the the bound
        given for p is only for the case were p does not divide
        xyz, i.e., p does not divide any of x, y, or z.
        
        Dan
445.9AUSSIE::GARSONFri Jun 19 1992 00:4836
re .7
    
>Also, in 1983 Faltings proved that for every n >= 3, fermat's equation has only
>finitely many primitive solutions (unlike the generators of Pythagorean
>triples).
    
    I happened to be reading about this the other day. Here are some more
    details that I couldn't find already mentioned in this conference.
    
    Suppose a^n + b^n = c^n then we can write this as x^n + y^n = 1 where
    x = a/c and y = b/c.
    
    Since for n=2 it is well known that there are an infinite number of
    primitive Pythagorean triples, the unit circle passes through
    infinitely many rational points.
    
    Mordell's Conjecture was that:
    
    No curve whose genus exceeds one can pass through more than a finite
    number of rational points.
    
    This was proved by the (then West) German mathematician Gerd Faltings
    for which he won a Fields Medal.
    
    Hence the result in .7 that for any given n there are no more than a
    finite number of primitive counter-examples to FLT. To my knowledge
    there is no known *upper* bound on "n" for counter-examples so
    therefore it is an open question as to whether the total number of
    primitive counter-examples to FLT is finite or infinite.

    (Implicit in the above is that the "supercircles" for n>2 are of genus
     greater than one. EFTR.)
    
    P.S. I am close to completing a proof of FLT but am temporarily out of
    1" by 11" paper. If nothing else I should be able to make a fairly
    large M�bius strip. :-)