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FLMTA Prize #1: S(n)=0 & n>(phi(x)*phi(y)*phi(z)/8) --> ten bucks
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DGoncz@aol.com
science forum Guru Wannabe


Joined: 25 Oct 2005
Posts: 122

PostPosted: Fri Jul 21, 2006 8:01 am    Post subject: Re: FLMTA Prize #1: S(n)=0 & n>(phi(x)*phi(y)*phi(z)/8) --> ten bucks Reply with quote

If n_min > lcm(phi(x), phi(y), phi(z)) / 2* gcd(phi(x),phi(y),phi(z)),

you win. Now that is a pecuiar expression.

The Dougster wrote:
Quote:
I think I can make it easier. I did a little more work

n_min > lcm( phi(x), phi(y), phi(z) ) / 4 takes the tenner.

Doug (who is still stuck at a limit of 41 for z, with x<y<z)
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DGoncz@aol.com
science forum Guru Wannabe


Joined: 25 Oct 2005
Posts: 122

PostPosted: Fri Jul 21, 2006 1:00 am    Post subject: Re: FLMTA Prize #1: S(n)=0 & n>(phi(x)*phi(y)*phi(z)/8) --> ten bucks Reply with quote

I think I can make it easier. I did a little more work

n_min > lcm( phi(x), phi(y), phi(z) ) / 4 takes the tenner.

Doug (who is still stuck at a limit of 41 for z, with x<y<z)
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DGoncz@aol.com
science forum Guru Wannabe


Joined: 25 Oct 2005
Posts: 122

PostPosted: Fri Jul 21, 2006 12:22 am    Post subject: Re: FLMTA Prize #1: S(n)=0 & n>(phi(x)*phi(y)*phi(z)/8) --> ten bucks Reply with quote

What a mess. I forget to say the only value for n that is of interest
is the *lowest* one. If any of you are inconvenienced by this, just
post the obvious, I'll fork over the tenner, and I'll work hard on
FLTMA Prize #2.

Sheez. I thought it was obvious, but I didn't say it!

The Dougster wrote:
Quote:
Hello, sci. math.

I have been tinkering with Fermat's Last Theorem and modular arithmetic
(FLTMA) for some time now.

I think I have found an upper limit for what I call the triple
coincindence.

If there were a counterexample (a,b,c,n) to FLT (a^n + b^n = c^n) ,
then by factoring gcd(a,b,c) there would be a primitive counterexample
x^n + y^n = z^n, but there won't be, because FLT is proved. There would
also be x^p + y^p = z^p but that is part of another problem.

Now for any x,y,z with gcd(x,y,z) = 1 and x<y<z<(x+y), there is a
series

S(n) with 0 <= n, defined as
(x^n + y^n) mod z +
(z^n - x^n) mod y +
(z^n - y^n) mod x

and this series isof interest to those interested in random number
generators, and a few interested in FLT.

Now, I think I have got an upper limit on n | S(n) = 0. It's
phi(x)*phi(y)*phi(z)/8. Why 8?

'Cause phi is always even for any non-small number, and there are three
terms, I guess... 2^3 = 8. Hell, I don't know. It just seems to work
out.

I've tested every x.y.z with gcd(x.y,z) =1 and x<y<z<(x+y) up to 41.

Maybe I missed one and you can score easy!

Run it up as high as you like....or can.

If you can prove that's not the right upper limit, or give a
counterexample, then good on you, hope you had fun, you'll be
recognized here when you post, and the first posted counterexample in
news:sci.math gets ten bucks from me by check, promptly, with my
thanks.

Since we'd have to all agree on a proof, I can't offer a prize for a
proof. Sorry. :(

Have fun with it! :)

Gawd, I hope I got it right when I wrote gcd = 1 instead of pairwise
coprime.... We'll see.

Doug Goncz
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DGoncz@aol.com
science forum Guru Wannabe


Joined: 25 Oct 2005
Posts: 122

PostPosted: Thu Jul 20, 2006 10:03 pm    Post subject: Re: FLMTA Prize #1: S(n)=0 & n>(phi(x)*phi(y)*phi(z)/8) --> ten bucks Reply with quote

Yes, phi is Euler's phi, not the golden section.

Doug
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Amadeus Train-Owwell Zirc
science forum Guru


Joined: 24 Mar 2005
Posts: 507

PostPosted: Thu Jul 20, 2006 9:57 pm    Post subject: Re: FLMTA Prize #1: S(n)=0 & n>(phi(x)*phi(y)*phi(z)/8) --> ten bucks Reply with quote

how is that a series?... phi is the golden section,
or the totient function?

Quote:
S(n) with 0 <= n, defined as
(x^n + y^n) mod z +
(z^n - x^n) mod y +
(z^n - y^n) mod x

and this series isof interest to those interested in random number
generators, and a few interested in FLT.

Now, I think I have got an upper limit on n | S(n) = 0. It's
phi(x)*phi(y)*phi(z)/8. Why 8?

'Cause phi is always even for any non-small number, and there are three
terms, I guess... 2^3 = 8. Hell, I don't know. It just seems to work
out.

I've tested every x.y.z with gcd(x.y,z) =1 and x<y<z<(x+y) up to 41.

Gawd, I hope I got it right when I wrote gcd = 1 instead of pairwise
coprime.... We'll see.

thus:
compression is only & always deployed around tension.

--it takes some to jitterbug!
http://members.tripod.com/~american_almanac
http://www.21stcenturysciencetech.com/2006_articles/Amplitude.W05.pdf
http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html
http://larouchepub.com/other/2006/3322_ethanol_no_science.html
http://www.wlym.com/pdf/iclc/howthenation.pdf
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DGoncz@aol.com
science forum Guru Wannabe


Joined: 25 Oct 2005
Posts: 122

PostPosted: Thu Jul 20, 2006 9:46 pm    Post subject: FLMTA Prize #1: S(n)=0 & n>(phi(x)*phi(y)*phi(z)/8) --> ten bucks Reply with quote

Hello, sci. math.

I have been tinkering with Fermat's Last Theorem and modular arithmetic
(FLTMA) for some time now.

I think I have found an upper limit for what I call the triple
coincindence.

If there were a counterexample (a,b,c,n) to FLT (a^n + b^n = c^n) ,
then by factoring gcd(a,b,c) there would be a primitive counterexample
x^n + y^n = z^n, but there won't be, because FLT is proved. There would
also be x^p + y^p = z^p but that is part of another problem.

Now for any x,y,z with gcd(x,y,z) = 1 and x<y<z<(x+y), there is a
series

S(n) with 0 <= n, defined as
(x^n + y^n) mod z +
(z^n - x^n) mod y +
(z^n - y^n) mod x

and this series isof interest to those interested in random number
generators, and a few interested in FLT.

Now, I think I have got an upper limit on n | S(n) = 0. It's
phi(x)*phi(y)*phi(z)/8. Why 8?

'Cause phi is always even for any non-small number, and there are three
terms, I guess... 2^3 = 8. Hell, I don't know. It just seems to work
out.

I've tested every x.y.z with gcd(x.y,z) =1 and x<y<z<(x+y) up to 41.

Maybe I missed one and you can score easy!

Run it up as high as you like....or can.

If you can prove that's not the right upper limit, or give a
counterexample, then good on you, hope you had fun, you'll be
recognized here when you post, and the first posted counterexample in
news:sci.math gets ten bucks from me by check, promptly, with my
thanks.

Since we'd have to all agree on a proof, I can't offer a prize for a
proof. Sorry. :(

Have fun with it! :)

Gawd, I hope I got it right when I wrote gcd = 1 instead of pairwise
coprime.... We'll see.

Doug Goncz
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