T.R | Title | User | Personal Name | Date | Lines |
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445.1 | First look at what CAN be proved | METOO::YARBROUGH | | Mon Mar 03 1986 10:01 | 12 |
| 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.
|
445.2 | Still bigger lower limits... | 26608::YARBROUGH | Why is computing so labor intensive? | Thu May 05 1988 15:16 | 6 |
| 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!
|
445.3 | I'll take the setup | POOL::HALLYB | You have the right to remain silent. | Thu May 05 1988 15:21 | 5 |
| > 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
|
445.4 | I got a good chuckle out of this, too | VMSDEV::HALLYB | The Smart Money was on Goliath | Fri Nov 03 1989 09:00 | 21 |
| 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
|
445.5 | for all practical purposes ;-) | CIVAGE::LYNN | Lynn Yarbrough @WNP DTN 427-5663 | Mon Jan 13 1992 15:32 | 7 |
| >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.
|
445.7 | | TRACE::GILBERT | Ownership Obligates | Wed May 06 1992 12:28 | 10 |
| 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).
|
445.8 | | GUESS::DERAMO | Dan D'Eramo, zfc::deramo | Wed May 06 1992 12:44 | 7 |
| 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
|
445.9 | | AUSSIE::GARSON | | Fri Jun 19 1992 00:48 | 36 |
| 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. :-)
|