Re: Fermat's Last Theorem 'Near Misses'
Arthur J. O'Dwyer
On Mon, 14 Jun 2004, Gary Edstrom wrote:
>
> Of course we all know that FLT has supposedly been proven (Although the
> number theory involved is well beyond my capability to follow).
>
> However, there are some impressive 'near misses'.
>
> Consider the cubes of 64 and 94:
>
> 64^3 + 94^3 - 103^3 = 1
>
> Can anyone come up with even more impressive 'near misses'?
There was reportedly one in one of the Simpsons Halloween episodes.
Google google...
1782^12 + 1841^12 = 1922^12
Well, my calculator says it's correct...
-Arthur,
an ingenious parity
Keith A. Lewis
>> Can anyone come up with even more impressive 'near misses'?
>
> There was reportedly one in one of the Simpsons Halloween episodes.
>Google google...
>
> 1782^12 + 1841^12 = 1922^12
>
>Well, my calculator says it's correct...
Did the Simpsons really discuss Fermat, or was it drawn on a chalkboard or
something?
My calc program says both sides are 2.54121026e+39, but the difference is
7.0021195e+29.
It's easy to see that it's incorrect; even + odd = odd.
--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
Mark Brader
> > There was reportedly one in one of the Simpsons Halloween episodes.
Keith Lewis:
> My calc program says both sides are 2.54121026e+39, but the difference is
> 7.0021195e+29.
Or more precisely:
1,782^12 = 1,025,397,835,622,633,634,807,550,462,948,226,174,976
1,841^12 = 1,515,812,422,991,955,541,481,119,495,194,202,351,681
total = 2,541,210,258,614,589,176,288,669,958,142,428,526,657
1,922^12 = 2,541,210,259,314,801,410,819,278,649,643,651,567,616
difference = 700,212,234,530,608,691,501,223,040,959
--
Mark Brader, Toronto "More importantly, Mark is just plain wrong."
m...@vex.net -- John Hollingsworth
Phil Carmody
> Of course we all know that FLT has supposedly been proven (Although the
> number theory involved is well beyond my capability to follow).
>
> However, there are some impressive 'near misses'.
>
> Consider the cubes of 64 and 94:
>
> 64^3 + 94^3 - 103^3 = 1
>
> Can anyone come up with even more impressive 'near misses'?
>
> Either higher powers or larger numbers, but the sum should differ by
> only 1 from a perfect cube.
Quoth Dr. David Broadhurst, only yesterday:
<<<
[How?]
By trying to understand sections 5.6 -- 5.10 of Cohen's book,
A Course in Computational Algebraic Number Theory.
I just got another hit with U/V(p,q,n), q^2=1 and small p.
This time [p,q]=[91,1], but the taxi-cab seed took several
CPU days to find:
664572^3 + 257118^3 == 677161^3 + (-1)^3 =~ 3.1*10^17
>>>
(that's part of a thread on the 'primeform' yahoogroup, a group
for people whose interest lies in finding large, interesting,
or otherwise curious prime numbers (using the 'primeform' program
usually))
Other smaller ones are
9^3 + (-8)^3 = 217 = 6^3 + 1
10^3 + 9^3 = 1729 = 12^3 + 1 (Ramanujan's TaxiCab number)
144^3 + (-71)^3 = 172^3 + (-135)^3 = 2628073 = 138^3 + 1
150^3 + (-73)^3 = 138^3 + 71^3 = 2985983 = 144^4 -1
...
Some other numbers expresible as x^3+y^3 and z^3+1 with x,y,z>0
(generated from a very quick and probably buggy script, these are worth
double-checking, and are probably not complete):
3375001
15438250
121287376
401947273
3680797185
6352182209
7856862273
12422690497
73244501505
145697644729
179406144001
648787169394
938601300672
985966166178
1594232306569
....
Phil
--
1st bug in MS win2k source code found after 20 minutes: scanline.cpp
2nd and 3rd bug found after 10 more minutes: gethost.c
Both non-exploitable. (The 2nd/3rd ones might be, depending on the CRTL)
Nir
Your calculator doesn't have enough digits...
if 1782^(12) + 1841^(12) = 1922^(12), it mean that
1782^12+1841^12-1922^12=0, but it's not, it's equal to -7.0021^(29)...
Dan Tilque
> Other smaller ones are
>
> 9^3 + (-8)^3 = 217 = 6^3 + 1
Which looks better if rewritten as 6^3 + 8^3 = 9^3 - 1
Not counting trivial ones (those involving only 0 and 1 as base
numbers), this seems to be the smallest near miss.
--
Dan Tilque
Don Del Grande
> There was reportedly one in one of the Simpsons Halloween episodes.
>Google google...
>
> 1782^12 + 1841^12 = 1922^12
It was in "Homer^3" as one of the things Homer saw in the 3-D
universe.
No need for working it all out to show that it is wrong; 1782^12 is an
even number, 1841^12 is odd, and 1922^12 is even, so you would have
even + odd = even.
-- Don
Ted S.
wrote in news:Pine.LNX.4.58-035....@unix48.andrew.cmu.edu:
> It floated past Homer^3 when he fell into... THE THIRD DIMENSION!
> (woooo weeeooooo...) There were other bizarre objects in the Third
> Dimension, but I don't remember what they were. Surely *somebody*
> on rec.puzzles must have the entire Simpsons oeuvre on DVD...?
Why don't you just go to snpp.com and look up the relevant THOH episode?
:-)
And no, nobody has the entire Simpsons śuvre on DVD because the relevant
DVDs haven't been released yet, a subject of much consternation on
alt.tv.simpsons.... :-)
One of the things in the Third Dimension was a hexadecimal code for "FRINK
RULES"
--
Ted <fedya at bestweb dot net>
The way I see it, you raised three children who could knock out and hog-
tie a perfect stranger, you must be doing *something* right.
Marge Simpson, <http://www.snpp.com/episodes/7G01.html>
Phil Carmody
Indeed it is (at least for cubes, that is).
I rewote my script, and now have what I believe to be a complete
list of all near misses less than or equal to 43451786041259765624
= 3515550^3 + 140624^3 = 3515625^3 - 1.
I have 108 x^3+y^3 = z^3+1, 1<y<x<z<10^7, 30 of which are parametrisable
as (9*b^4)^3 + (9*b^3+1)^3 = (9*b^4+3*b)^3 + 1
and 97 x^3+y^3 = z^3-1, 1<y<x<z<=3515625, which I've not tried to
parametrise yet, but I'm sure that there will be at least one simple
pattern I can find.
According to Dr. Broadhurst, almost all solutions with fewer than
billions of digits lie on one of about 7 triply-parametrised surfaces.
I may have misunderstood him, but I suspect that one of my data
points may provide an 8th such surface. However, there are many
single points with small magnitudes which lie on surfaces whose
parametrisations mean that all other points on it would have billions
of digits.
If anyone has an spare CPU sitting around (Athlon64 in particular, as
GMP is fastest on such machines), and wishes to hunt for further
solutions for a couple of days or so, please let me know, and I'll
send you my GP/Pari script. Maybe a 9th small paramterisation can be
found!
Zdenek Jizba
Here are some of the integers z that satisfy the near cubes version
discussed here. At the time nobody came up with the solution
x^3 + y^3 = z^3 +/- 1
Following is the text of my 2001 problem posed to this group:
One month ago (i.e 11/8/2001) I submitted a problem that has not been
solved by anyone in this newsgroup. Frankly I was not surprised because
it is a rather difficult one. I have no idea whether the set of integers
is finite or infinite. It would be in my opinion extremely difficult
to prove it one way or another, and besides it may not be worth the
effort. I am quite certain that for any integer in this set there is no
way to predict it from a subset. I have sorted the elements of this
set in an increasing sequence, but this has no significance.
Today I am adding two more clues. The original problem follows:
"Here is an interesting one. The numbers below have a common property
9 12 103 144 150 172 249 . . . .
Find out what that property is. Here are the clues:
1. The answer is in the domain of the Queen of Mathematics
2. Some people would call the header a clue"
Agent Maxwel Smart would move his index fingers together
and say "it's THIS close".
New clues:
3. The set can also be represented as two distinct subsets
9, 144, 172, 505, 577, 729, ...
12, 103, 150, 249, 495, 738, ....
4. Each subset of clue 3 is associated with a specific
diophantine equation.
Richard Sabey
wrote in message news:87acz4h...@nonospaz.fatphil.org:
> "Dan Tilque" <dti...@nwlink.com> writes:
>> Phil Carmody wrote:
>>
>> > Other smaller ones are
>> >
>> > 9^3 + (-8)^3 = 217 = 6^3 + 1
>>
>> Which looks better if rewritten as 6^3 + 8^3 = 9^3 - 1
> I rewote my script, and now have what I believe to be a complete
> list of all near misses less than or equal to 43451786041259765624
> = 3515550^3 + 140624^3 = 3515625^3 - 1.
>
> I have 108 x^3+y^3 = z^3+1, 1<y<x<z<10^7, 30 of which are parametrisable
> as (9*b^4)^3 + (9*b^3+1)^3 = (9*b^4+3*b)^3 + 1
> and 97 x^3+y^3 = z^3-1, 1<y<x<z<=3515625, which I've not tried to
> parametrise yet, but I'm sure that there will be at least one simple
> pattern I can find.
There is the pattern
x = 9b^4-3b
y = 9bł-1
z = 9b^4
=> xł + ył = zł - 1
identical to the pattern you found except for changes of sign and
for rearrangements such as the one suggested by Dan Tilque.
b=1 and b=25 give the examples you quoted above.
--
Richard Sabey Visit the r.p.crosswords competition website
cryptic_fan at hotmail.com http://www.rsabey.pwp.blueyonder.co.uk/rpc/
Phil Carmody
Indeed. This sequence is but a one-dimensional subset of Dr. Broadhurst's
[p,q]=[432*t^6-2,1] Lucas-based triple-parametrisation. Ramanujan's being
the smallest one on the p=430 curve.
I'm reliably informed by the good doctor that I've just discovered a
new family that generates sequences with fewer than a trillion digits
(the previous billion was a Big-Oh underestimate),
> 2174679^3 + 270036^3 = 2176066^3 - 1 = 10304245329684255495
[p,q]=[202751082679598,1]
Hence the call for CPU volunteers, to find examples on curves with
"small regulators" (over my head, I'm afraid), which give exponential
sequences that don't bloat to trillions of digits at their second
term. (My poor P3/700 laptop is so hot I can't even press Ctrl-Z to
let it cool down!)
Gerry Myerson
Phil Carmody <thefatphi...@yahoo.co.uk> wrote:
> Cross-posting to sci.math due to call for CPU power...
>
> "Dan Tilque" <dti...@nwlink.com> writes:
> > Phil Carmody wrote:
> >
> > > Other smaller ones are
> > >
> > > 9^3 + (-8)^3 = 217 = 6^3 + 1
> >
> > Which looks better if rewritten as 6^3 + 8^3 = 9^3 - 1
> >
> > Not counting trivial ones (those involving only 0 and 1 as base
> > numbers), this seems to be the smallest near miss.
>
> Indeed it is (at least for cubes, that is).
>
> I rewote my script, and now have what I believe to be a complete
> list of all near misses less than or equal to 43451786041259765624
> = 3515550^3 + 140624^3 = 3515625^3 - 1.
Have you seen Noam Elkies' page on Fermat Near-Misses,
http://www.math.harvard.edu/~elkies/ferm.html
Elkies also mentions Dan Bernstein's tables,
http://cr.yp.to/threecubes.html
and Hisanori Mishima's page,
http://www.asahi-net.or.jp/~KC2H-MSM/mathland/math04/cube01.htm
Maybe these were already noted in rec.puzzles - I've only seen
the sci.math part of the thread.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Phil Carmody
Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> writes:
> > I rewote my script, and now have what I believe to be a complete
> > list of all near misses less than or equal to 43451786041259765624
> > = 3515550^3 + 140624^3 = 3515625^3 - 1.
>
> Have you seen Noam Elkies' page on Fermat Near-Misses,
> http://www.math.harvard.edu/~elkies/ferm.html
> Elkies also mentions Dan Bernstein's tables,
> http://cr.yp.to/threecubes.html
> and Hisanori Mishima's page,
> http://www.asahi-net.or.jp/~KC2H-MSM/mathland/math04/cube01.htm
Thanks Gerry.
I was aware of the good professors' involvement with the subject, and
had already investigated Bernstein's 'sortedsums' and associated code.
The DJB reference you give alas has a "a^3+b^3+c^3 is not a cube" rider,
which precludes +/-1. It appears that the Elkies paper,
"Rational points near curves and small nonzero |x^3-y^2| via lattice reduction"
http://arxiv.org/abs/math/0005139
may be useful.
Alas, I have no experience with lattice reduction (I know roughly what
it's supposed to achieve, but don't know how it achieves it, or even
how to play with it in order to learn (using GP/Pari) through examples),
so it might be tough going.