Soln found to a^5+b^5+c^5 = d^5+e^5+f^5, with a+b+c = d+e+f = 0
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Newsgroups: sci.math, sci.math.symbolic
From:
TPiezas <tpie... @gmail.com> Date: Tue, 3 Nov 2009 03:12:22 -0800 (PST)
Local: Tues, Nov 3 2009 10:12 pm
Subject: Soln found to a^5+b^5+c^5 = d^5+e^5+f^5, with a+b+c = d+e+f = 0
Hello all, I asked this question about fifth powers before, and turns out the
i) a^k+b^k+c^k = d^k+e^k+f^k, with a+b+c = d+e+f = 0, for k = 1,5,
have non-trivial solns in the integers. The case (i) was found by
Then there's always 9th powers....
See: http://sites.google.com/site/tpiezas/020
- Titus
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From:
Gordon Stangler <gordon.stang... @gmail.com> Date: Tue, 3 Nov 2009 14:06:50 -0800 (PST)
Local: Wed, Nov 4 2009 9:06 am
Subject: Re: Soln found to a^5+b^5+c^5 = d^5+e^5+f^5, with a+b+c = d+e+f = 0
On Nov 3, 5:12 am, TPiezas <tpie... @gmail.com> wrote:
> Hello all,
> I asked this question about fifth powers before, and turns out the
> i) a^k+b^k+c^k = d^k+e^k+f^k, with a+b+c = d+e+f = 0, for k = 1,5,
> have non-trivial solns in the integers. The case (i) was found by
> Then there's always 9th powers....
> See:http://sites.google.com/site/tpiezas/020
> - Titus
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From:
"alainvergh... @gmail.com" <alainvergh... @gmail.com> Date: Wed, 4 Nov 2009 02:04:52 -0800 (PST)
Local: Wed, Nov 4 2009 9:04 pm
Subject: Re: Soln found to a^5+b^5+c^5 = d^5+e^5+f^5, with a+b+c = d+e+f = 0
On 3 nov, 23:06, Gordon Stangler <gordon.stang... @gmail.com> wrote:
> On Nov 3, 5:12 am, TPiezas <tpie
... @gmail.com> wrote:
> > Hello all,
> > I asked this question about fifth powers before, and turns out the
> > i) a^k+b^k+c^k = d^k+e^k+f^k, with a+b+c = d+e+f = 0, for k = 1,5,
> > have non-trivial solns in the integers. The case (i) was found by
> > Then there's always 9th powers....
> > See:http://sites.google.com/site/tpiezas/020
> > - Titus
> Nice site! It has lots of useful information.- Masquer le texte des messages précédents -
> - Afficher le texte des messages précédents -
I do share the pleasure of Titus,
Alain
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Newsgroups: sci.math, sci.math.symbolic
From:
TPiezas <tpie... @gmail.com> Date: Wed, 4 Nov 2009 06:54:27 -0800 (PST)
Local: Thurs, Nov 5 2009 1:54 am
Subject: Re: Soln found to a^5+b^5+c^5 = d^5+e^5+f^5, with a+b+c = d+e+f = 0
On Nov 4, 4:04 am, "alainvergh... @gmail.com" <alainvergh... @gmail.com>
> On 3 nov, 23:06, Gordon Stangler <gordon.stang
... @gmail.com> wrote:
> > On Nov 3, 5:12 am, TPiezas <tpie... @gmail.com> wrote:
> > > Hello all,
> > > I asked this question about fifth powers before, and turns out the
> > > i) a^k+b^k+c^k = d^k+e^k+f^k, with a+b+c = d+e+f = 0, for k = 1,5,
> > > have non-trivial solns in the integers. The case (i) was found by
> > > Then there's always 9th powers....
> > > See:http://sites.google.com/site/tpiezas/020
> > > - Titus
> > Nice site! It has lots of useful information.- Masquer le texte des messages précédents -
> > - Afficher le texte des messages précédents -
> Welcome to both of you,
> I do share the pleasure of Titus,
> Alain- Hide quoted text -
> - Show quoted text -
(x+z)^5 + (-x+z)^5 + (y+z)^5 + (-y+z)^5 + (x+y+z)^5 + (-x-y+z)^5 =
where x^2+xy+y^2 = z^2. However, if we choose to partition 66 as a
(x+z)^k + (-x+z)^k + (y+z)^k + (-y+z)^k + (x+y+z)^k + (-x-y+z)^k = 2
suddenly good for k = 1,2,3,4,5. Anybody knows why? Or if there is 7th
For ex, choose the smallest soln {x,y,z} = {3,5,7}. Then we get,
10^5+4^5+12^5+2^5+15^5+(-1)^5 = 66*7^5
or, alternatively,
10^k+4^k+12^k+2^k+15^k+(-1)^k = 2*7^k + 2*(2*7)^k, for k = 1,2,3,4,5.
Just one of the many surprises about identities. For more, see here: http://sites.google.com/site/tpiezas/022
- Titus
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