New Discovery

Turtle · 07-30-2005

http://hypography.com/forums/newthre...=newthread&f=4

Spiders welcome. Mind my web.

Turtle · 07-30-2005

___This test is already very interesting. I note I posted at 1:45 AM PDST in order to test the search engines, spiders, scavengers, views etc. There was no New Discovery at 1:45 AM that I knew of; my interest lay in the phrase.
___Later, a little before 3:00 AM PDST I went to my regular check of spaceweather.com to see how the new X class flare was developing & the site had updated the main page with the New Tenth Planet Discovery. I then started the thread with the announcement at post time 3:00AM PDST.
___More evidence for telepathy? Random quantum flucuation? Serendipity? Word play?
___Very interesting so far. Now is the time when we stand on our heads against the wall; weak necked folks use the corners.

Turtle · 08-14-2005

Hi Buffy. Hi Tormod.

PS I do actually have a new discovery today. A tetrahedral proof that no solutions exist for X^3 + Y^3 =Z^3

Katabatak Math-An Exploration In Pure Number Theory
It made need some algebraic shaping up to satisfy a nit picker; it is patently geometrically obvious to me. Amazing.

C1ay · 08-15-2005

Quote:

Originally Posted by Turtle

Hi Buffy. Hi Tormod.

PS I do actually have a new discovery today. A tetrahedral proof that no solutions exist for X^3 + Y^3 =Z^3

Katabatak Math-An Exploration In Pure Number Theory
It made need some algebraic shaping up to satisfy a nit picker; it is patently geometrically obvious to me. Amazing.

Sorry but it's not new. Fermat's Last Theorem stated that there are no integer solutions for x^n+y^n=z^n for n>2. Euler proved the case for n=3.

Tormod · 08-15-2005

Maybe Turtle stumbled upon a simple geometric solution? Would be nice!

Turtle · 08-15-2005

Quote:

Originally Posted by Tormod

Maybe Turtle stumbled upon a simple geometric solution? Would be nice!

___That's what I meant; a new different discovery. The sum of cubes is a specific example of Fermat's general case. Euler may have proved it for cubes, but not like this. Wiles proved it for all cases of powers over two, but I continue with my Katabatak look into it anyway. That something is done once, is no reason not to find another way to do it.
___Fuller may even comment somewhere in the text on it; I gazed at the drawing for a few hours & in an ephiphany saw it quite clear.

Turtle · 08-15-2005

___No wonder we encountered some confusion; I linked to the wrong drawing in post 111 of the Katabatak thread!

___If you please to review it now as I have made corrections, all that I said about it still holds. Science is always amendable.

http://hypography.com/forums/showthr...4772#post54772

___How simply palindromic! Post 111 of the Katabatak. Chaos favors the prepared imagination.

Turtle · 08-21-2005

___Rather than continue this in the Katabatak thread where it is slightly off topic now, I decided to carry on here.

____Eureka! A little epiphany. The problem at hand is in regard to this Fuller drawing:
http://www.rwgrayprojects.com/synerg...igs/f9001.html

___Observing on the drawing the lower right hand quadrant & the stacking of tetrahedrons that form powers of 3. Although the geometry is clearly a proof to me that no two powers of three sum to a power of three, I expressed a desire for a formula for deriving the number of tetrahedrons in each successive layer, i.e. a formula where n is the layer number (frequency in close-packing terms) & the result is how many tetrahedrons in that layer. Fuller gives the starting as 1, 7, 19, 37... I noted they differed by multiples of 6 & the question is which multiples. The first is 1 [1*6], the second is 3 [3*6], then 6, then 10 & so on. Does the se {1,3,6,10...} look familiar?
___It is the set of triangular numbers! So now I have my formula; it is the formula for triangular number, times 6, plus 1.

___Therfore, for n = the layer, the number of close-packed tetrahedrons is
(n/2 * (n+1)) * 6 + 1

___Anyway, it's the middle of the night here & I jumped up from near sleep to post this; I intend to take it up again.