A Cojecture 
Author Message
 A Cojecture

 In alt.math.recreational I introduced the following conjecture which
if true, would be a generalization of Fermat's last theorem.

 FOR N GREATER THAN 3, THERE ARE NO DISTINCT
   N+1 INTEGERS SUCH THAT THE LARGEST ONE RAISED
  TO THE POWER N EQUALS TO THE SUM OF THE REMAINING
 ONES RAISED TO THE POWER N.

  I tested this conjecture for N=4 up to integer 25, for N=5  up to
integer
14 and for N=6 up to integer 9. My APL2 function is crude. It merely
generates outer products of the sums of all possible powers of N, and
if any one or more sums equals (N+1)*N, then it prints a vector
(actually
all permutations of the vector) containing the integers. The function
has
produced many results for N=3 including some quite interesting ones.

 My challenge to the APL community is to write a more efficient APL
function and possibly find a counter example of the conjecture.



Sat, 27 Mar 2004 15:59:54 GMT  
 A Cojecture


Quote:
> In alt.math.recreational I introduced the following conjecture which
>if true, would be a generalization of Fermat's last theorem.

> FOR N GREATER THAN 3, THERE ARE NO DISTINCT
>   N+1 INTEGERS SUCH THAT THE LARGEST ONE RAISED
>  TO THE POWER N EQUALS TO THE SUM OF THE REMAINING
> ONES RAISED TO THE POWER N.

...

Unfortunately this is not true, even if "N+1" is replaced by "N"
(i.e. one of the integers is 0).  If you want to look for a solution
numerically using APL, then I think you'll need a bigger workspace!
Spoiler (in J) follows after suitable spoiler space:
 L
 o
 o
 k/
 u
 p/
 R
 o
 g
 e
 r/
 F
 r
 y
 e,/
 N
 o
 a
 m/
 E
 l
 k
 i
 e
 s/
 o
 r/
 "E
 u
 l
 e
 r'
 s/
 S
 p
 o
 i
 l
 e
 r
 s".

+`+`-/ 95800 217519 414560 422481 ^ 4x      NB. smallest counterexample
--

  Department of Statistics,  University of Warwick,  Coventry CV4 7AL,  U.K.
  http://www.warwick.ac.uk/statsdept/Staff/JEHS/



Sat, 27 Mar 2004 18:07:24 GMT  
 A Cojecture
My apologies. The following lines are not clear

Quote:
>  My APL2 function is crude. It merely
> generates outer products of the sums of all possible powers of N, and
> if any one or more sums equals (N+1)*N, then it prints a vector...

Replace these lines by:

   My APL2 function is crude. It merely generates outer products of the
sums of all possible powers of N. If any one or more sums equals n*N
where n is the largest of the N+1 integers, then it prints a vector...

(Also I should have pointed out that with -ws 30m parameters the
 function produces a WS FULL message at the given values of n
 25, 14 and 9 listed above)



Sat, 27 Mar 2004 23:55:26 GMT  
 A Cojecture
 Your answers do not make sense. Even if the numbers exceed the limit
for APL2 integers, my tests below do not indicate that your numbers
prove the conjecture false

  (95800 217519 414560 422481)*.25
17.59305973 21.59604877 25.37446235 25.49481093
      +/95800 217519 414560 422481*4
6.371749968E22
      6.371749968E22*.25
502417.4112
      502417*4
6.371729111E22
      502418*4
6.371779839E22

Quote:



> > In alt.math.recreational I introduced the following conjecture which
> >if true, would be a generalization of Fermat's last theorem.

> > FOR N GREATER THAN 3, THERE ARE NO DISTINCT
> >   N+1 INTEGERS SUCH THAT THE LARGEST ONE RAISED
> >  TO THE POWER N EQUALS TO THE SUM OF THE REMAINING
> > ONES RAISED TO THE POWER N.

> ...

> Unfortunately this is not true, even if "N+1" is replaced by "N"
> (i.e. one of the integers is 0).  If you want to look for a solution
> numerically using APL, then I think you'll need a bigger workspace!
> Spoiler (in J) follows after suitable spoiler space:
>  L
>  o
>  o
>  k/
>  u
>  p/
>  R
>  o
>  g
>  e
>  r/
>  F
>  r
>  y
>  e,/
>  N
>  o
>  a
>  m/
>  E
>  l
>  k
>  i
>  e
>  s/
>  o
>  r/
>  "E
>  u
>  l
>  e
>  r'
>  s/
>  S
>  p
>  o
>  i
>  l
>  e
>  r
>  s".

> +`+`-/ 95800 217519 414560 422481 ^ 4x      NB. smallest counterexample
> --

>   Department of Statistics,  University of Warwick,  Coventry CV4 7AL,  U.K.
>   http://www.warwick.ac.uk/statsdept/Staff/JEHS/




Sun, 28 Mar 2004 00:13:47 GMT  
 A Cojecture

Quote:

>  Your answers do not make sense. Even if the numbers exceed the limit
> for APL2 integers, my tests below do not indicate that your numbers
> prove the conjecture false

Shaw's answer made sense to me.

I believe his point was that  (0, 95800, 217519, 414560)
all raised to the fourth power add to 422481 raised to the fourth
thus proving your conjecture false.
Another way of saying this in J is:
 +/0 95800 217519 414560 ^ 4x      NB. smallest counterexample
31858749840007945920321
 422481 ^ 4x      NB. smallest counterexample
31858749840007945920321



Sun, 28 Mar 2004 00:59:04 GMT  
 A Cojecture
 I did not realize that zero is an integer!
Quote:


> >  Your answers do not make sense. Even if the numbers exceed the limit
> > for APL2 integers, my tests below do not indicate that your numbers
> > prove the conjecture false

> Shaw's answer made sense to me.

> I believe his point was that  (0, 95800, 217519, 414560)
> all raised to the fourth power add to 422481 raised to the fourth
> thus proving your conjecture false.
> Another way of saying this in J is:
>  +/0 95800 217519 414560 ^ 4x      NB. smallest counterexample
> 31858749840007945920321
>  422481 ^ 4x      NB. smallest counterexample
> 31858749840007945920321



Sun, 28 Mar 2004 01:13:54 GMT  
 A Cojecture

Quote:

>  I did not realize that zero is an integer!

It surely is.


Sun, 28 Mar 2004 01:23:45 GMT  
 A Cojecture
You should be able to reproduce that following J results
in APL2:

   +/ 95800 217519 414560  ^ 4
3.18587e22
   422481 ^ 4
3.18587e22

Same results displayed to 23 decimal places:

   23 ": +/ 95800 217519 414560 ^ 4
31858749840007946000000
   23 ": 422481 ^ 4
31858749840007946000000

Using extended precision numbers in J:

   +/ 95800 217519 414560 ^ 4x
31858749840007945920321
   422481 ^ 4x
31858749840007945920321

Quote:
----- Original Message -----


Sent: Tuesday, October 09, 2001 10:14 AM
Subject: Re: A Conjecture


>  Your answers do not make sense. Even if the numbers exceed the limit
> for APL2 integers, my tests below do not indicate that your numbers
> prove the conjecture false

>   (95800 217519 414560 422481)*.25
> 17.59305973 21.59604877 25.37446235 25.49481093
>       +/95800 217519 414560 422481*4
> 6.371749968E22
>       6.371749968E22*.25
> 502417.4112
>       502417*4
> 6.371729111E22
>       502418*4
> 6.371779839E22



Sun, 28 Mar 2004 01:54:26 GMT  
 A Cojecture
 I have no doubt that a counter-example to the conjecture exists
(see my comments in alt.math.recreational). I do object, however
to the inclusion of zero as one of the valid integers in the
conjecture. In fact it is very easy for example to find cubes that
are sums of two squares (25 and 100 come to mind). My note
in alt.math.recreational does NOT consider such cases.
Quote:

> You should be able to reproduce that following J results
> in APL2:

>    +/ 95800 217519 414560  ^ 4
> 3.18587e22
>    422481 ^ 4
> 3.18587e22

> Same results displayed to 23 decimal places:

>    23 ": +/ 95800 217519 414560 ^ 4
> 31858749840007946000000
>    23 ": 422481 ^ 4
> 31858749840007946000000

> Using extended precision numbers in J:

>    +/ 95800 217519 414560 ^ 4x
> 31858749840007945920321
>    422481 ^ 4x
> 31858749840007945920321

> ----- Original Message -----


> Sent: Tuesday, October 09, 2001 10:14 AM
> Subject: Re: A Conjecture


> >  Your answers do not make sense. Even if the numbers exceed the limit
> > for APL2 integers, my tests below do not indicate that your numbers
> > prove the conjecture false

> >   (95800 217519 414560 422481)*.25
> > 17.59305973 21.59604877 25.37446235 25.49481093
> >       +/95800 217519 414560 422481*4
> > 6.371749968E22
> >       6.371749968E22*.25
> > 502417.4112
> >       502417*4
> > 6.371729111E22
> >       502418*4
> > 6.371779839E22



Sun, 28 Mar 2004 08:05:36 GMT  
 A Cojecture


Quote:
> I have no doubt that a counter-example to the conjecture exists
>(see my comments in alt.math.recreational). I do object, however
>to the inclusion of zero as one of the valid integers in the
>conjecture. In fact it is very easy for example to find cubes that
>are sums of two squares (25 and 100 come to mind). My note
>in alt.math.recreational does NOT consider such cases.

...

Sorry if you found my original comment obscure.
If you don't like 0 as an integer, then the following (in J)
verifies that the sum of the fourth powers of 30, 120, 272, & 31
is 353^4:

   +/30 120 272 315^4x
15527402881
   353^4x
15527402881

The web page  http://www.mathsoft.com/asolve/fermat/fermat.html
seems to suggest that this coincidence (mathematicians may prefer
the term "isolated result") can be found in Hardy & Wright's
"Theory of Numbers", which I don't possess.

Fourth powers modulo 16 are all either 0 or 1, showing that
one only need look for solutions of the form
(2a)^4 + (2b)^4 + (2c)^4 + (2d+1)^4 = (2e+1)^4,
but even with efficiencies like this a purely numerical search
for such formulae, particularly for higher powers, will need
a large workspace and a long time.

I couldn't view your original note since our newsreader claims
that "alt.math.recreational does not exist"---don't know why.

Here are a couple of other cute coincidences I found using J.
The first probably depends on deep properties of elliptic curves.
The second doesn't.

   17j14 ": ,.(^. _104 + 12 28 396^4) % %: 10 18 58  NB. deep result?
 3.14159590496940
 3.14159265631841
 3.14159265358979
   666 A. 'moaned'   NB. beastly but totally inconsequential result
daemon

Regards, Ewart Shaw.
--

  Department of Statistics,  University of Warwick,  Coventry CV4 7AL,  U.K.
  http://www.warwick.ac.uk/statsdept/Staff/JEHS/



Sun, 28 Mar 2004 22:53:18 GMT  
 A Cojecture
  Thanks for this information. It does not surprise me that there
is a counter-example. Also I feel that I owe an explanation for
the reason why I was so vehement about insisting on four non-zero
integers in the case of N=4. For the case N=2 (e.g. 3 square plus 4
square equal 5 square) there is a geometrical explanation. Squares
are two dimensional objects and two dimensions require two
orthogonal axes. For N=3 (e.g. 3 cube plus 4 cube plus 5 cube
equal 6 cube) there is also a "geometrical" interpretation. Cubes
are three dimensional object and therefore numerical properties
of cubes require three orthogonal axes. For N=4, I felt that a four
dimensional hypercube should have a "geometrical" interpretation
as well. From this point of view, I felt the sum of three integers raised
to the power 4 was a numerical "coincidence". The fact that I could not
find four low value integers for the case of N=4 suggested to me the
possiblity of the conjecture.
Quote:
> Sorry if you found my original comment obscure.
> If you don't like 0 as an integer, then the following (in J)
> verifies that the sum of the fourth powers of 30, 120, 272, & 31
> is 353^4:

>    +/30 120 272 315^4x
> 15527402881
>    353^4x
> 15527402881



Mon, 29 Mar 2004 06:11:09 GMT  
 
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