Turtle Posted July 30, 2005 Report Share Posted July 30, 2005 http://hypography.com/forums/newthread.php?do=newthread&f=4 Spiders welcome. Mind my web. Quote Link to comment Share on other sites More sharing options...
Turtle Posted July 30, 2005 Author Report Share Posted July 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. :) Quote Link to comment Share on other sites More sharing options...
Turtle Posted August 15, 2005 Author Report Share Posted August 15, 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 :) http://hypography.com/forums/showthread.php?t=1343&page=12&pp=10It made need some algebraic shaping up to satisfy a nit picker; it is patently geometrically obvious to me. Amazing. :lol: Quote Link to comment Share on other sites More sharing options...
C1ay Posted August 15, 2005 Report Share Posted August 15, 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 :) http://hypography.com/forums/showthread.php?t=1343&page=12&pp=10It made need some algebraic shaping up to satisfy a nit picker; it is patently geometrically obvious to me. Amazing. :lol: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. Quote Link to comment Share on other sites More sharing options...
Tormod Posted August 15, 2005 Report Share Posted August 15, 2005 Maybe Turtle stumbled upon a simple geometric solution? Would be nice! :lol: Quote Link to comment Share on other sites More sharing options...
Turtle Posted August 15, 2005 Author Report Share Posted August 15, 2005 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. :) Quote Link to comment Share on other sites More sharing options...
Turtle Posted August 15, 2005 Author Report Share Posted August 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/showthread.php?p=54772#post54772 ___How simply palindromic! Post 111 of the Katabatak. Chaos favors the prepared imagination. :evil: Quote Link to comment Share on other sites More sharing options...
Turtle Posted August 21, 2005 Author Report Share Posted August 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/synergetics/s09/figs/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. :P Quote Link to comment Share on other sites More sharing options...
Turtle Posted August 21, 2005 Author Report Share Posted August 21, 2005 ___Commercial break from the golf; must hurry. ;) ___Simplifying the formula we have: 3n^2+3n+1. Note: This formula is for close-packing tetrahedrons/spheres in/on a plane. ;) Quote Link to comment Share on other sites More sharing options...
Dark Mind Posted September 14, 2005 Report Share Posted September 14, 2005 Not sure. BUMP Dark Mind out :eek2:. Quote Link to comment Share on other sites More sharing options...
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