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From: TPiezas on 23 Jun 2010 13:54
Hello all,

Can anyone find another solution to,

a + b + c + d = x^2
a^2+b^2+c^2+d^2 = y^2
a^3+b^3+c^3+d^3 = z^3

other than {a,b,c,d} = {10, 13, 14, 44}?


- Titus
From: James Dow Allen on 23 Jun 2010 15:58
On Jun 24, 12:54 am, TPiezas <tpie...(a)gmail.com> wrote:
>
> a + b + c + d = x^2
> a^2+b^2+c^2+d^2 = y^2
> a^3+b^3+c^3+d^3 = z^3

Thanks: A problem that even a "brute" like me can tackle!

If (a, b, c, d) is a solution then so is
(akk, bkk, ckk, dkk)
for any square kk.

Solutions for a,b,c,d < 1300 with
no such common square factor include
(0, 0, 0, 1)
(10, 13, 14, 44)
(54, 109, 202, 260)
(102, 130, 234, 318)

James

From: spudnik on 23 Jun 2010 21:39
sic.

> > a + b + c + d = x^2
> > a^2+b^2+c^2+d^2 = y^2
> > a^3+b^3+c^3+d^3 = z^3

> If (a, b, c, d) is a solution then so is
>    (akk, bkk, ckk, dkk)
> for any square kk.
>
> Solutions for a,b,c,d < 1300 with
> no such common square factor include
>    (0, 0, 0, 1)
>    (10, 13, 14, 44)
>    (54, 109, 202, 260)
>    (102, 130, 234, 318)

thus&so:
surely it could not be so hard,
to find some of the rather definitive un-null results
of Michelson, Morely et al; is it?... well, even
as Albert the Witnit wobbled on the idea of aether,
it is really a matter of interpretation. so,
why cannot the electromagnetic properties
of atoms in "space" be an aether; to wit,
permitivity & permeability?

should your "theory" can be taken at all seriously,
you'd have to be able to explain such; would you not?

oh, and there never was a twin paradox;
it is just a "term of art" and pop-science. I mean,
shouldn't the few properties of energy, of light,
be of the ultimate importance for matter,
per the experiments of Young, Fresnel et al,
in utterly burying Newton's "theory" of corpuscles
-- til it was rescued by the word, "photon;
hereinat to be interpreted to mean a massless rock
o'light?... and, thanks for that Nobel!"

> Using Larmor’s transform, there is no twin’s paradox.

--BP loves Waxman-Obama cap&trade (at least circa Kyoto, or
Waxman's '91 cap&trade on NOX and SO2) --
how about a tiny tax, instead of the Last Bailout
of Wall Street and the "City of London?"
http://larouchepub.com/pr_lar/2010/lar_pac/100621pne_nordyke.html

--le theoreme prochaine du Fermatttt!
http://wlym.com
From: William Elliot on 24 Jun 2010 00:49
On Wed, 23 Jun 2010, TPiezas wrote:

> Hello all,
>
> Can anyone find another solution to,
>
> a + b + c + d = x^2
> a^2 + b^2 + c^2 + d^2 = y^2
> a^3 + b^3 + c^3 + d^3 = z^3
>
> other than {a,b,c,d} = {10, 13, 14, 44}?
>
{ 0,0,0,n^2 }

----
From: TPiezas on 24 Jun 2010 05:30
On Jun 23, 1:58 pm, James Dow Allen <jdallen2...(a)yahoo.com> wrote:
> On Jun 24, 12:54 am, TPiezas <tpie...(a)gmail.com> wrote:
>
>
>
> > a + b + c + d = x^2
> > a^2+b^2+c^2+d^2 = y^2
> > a^3+b^3+c^3+d^3 = z^3
>
> Thanks: A problem that even a "brute" like me can tackle!
>
> If (a, b, c, d) is a solution then so is
>    (akk, bkk, ckk, dkk)
> for any square kk.
>
> Solutions for a,b,c,d < 1300 with
> no such common square factor include
>    (0, 0, 0, 1)
>    (10, 13, 14, 44)
>    (54, 109, 202, 260)
>    (102, 130, 234, 318)
>
> James


Thanks, James! There are in fact, an infinite number of distinct
solutions with no common square factors, as the problem can be
approached via an elliptic curve, but the numbers grow so fast, and
may involve negative integers such as this one,

{a,b,c,d} = {-19925019218, 74350015661, 62412666610, 46437326276}

and I wanted to know to know if there were relatively small solns in
the positive integers.

P.S. Your answer has been added to the section "Martin quadruplets" at
http://sites.google.com/site/tpiezas/updates06

- Titus
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