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 March 23rd, 2019, 04:19 AM #1 Member   Joined: Aug 2015 From: Chiddingfold, Surrey Posts: 57 Thanks: 3 Math Focus: Number theory, Applied maths Fermat's last theorem Have I found a novel way of expressing Fermat's last theorem as follows? N = nt{ K.A^(3-p)} where N is the number of primitive triples a, b, c and a^p = b^p +c^p and a all values up to A. (K is about .155, 0.152929 for p =1 and 0.159155 for p = 2) For example, this gives for a up to 10,000, P = 1 N = 15292928 (actually 15198740) and for a up to 1000 p = 2 N = 318 (actually 319) p > 2 N = 0 with any values for A and K apart from infinity. Last edited by skipjack; March 29th, 2019 at 06:14 PM. March 23rd, 2019, 08:09 AM   #2
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 Originally Posted by magicterry Have I found a novel way of expressing Fermat's last theorem as follows? N =int{ K.A^(3-p)} where N is the number of primitive triples a, b, c and a^p = b^p +c^p and a all values up to A. (K is about .155, 0.152929 for p =1 and 0.159155 for p = 2) For example this gives for a up to 10,000, P = 1 N = 15292928 (actually 15198740) and for a up to 1000 p =2 N = 318 (actually 319) p > 2 N = 0 with any values for A and K apart from infinity. What are N, K, and A supposed to be?

-Dan

Last edited by skipjack; March 29th, 2019 at 06:14 PM. March 23rd, 2019, 01:34 PM   #3
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 Originally Posted by topsquark What are N, K, and A supposed to be? -Dan
Thou shalt find them 3 critters in the alphabet... March 28th, 2019, 03:43 AM   #4
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Quote:
 Originally Posted by magicterry Have I found a novel way of expressing Fermat's last theorem as follows? N = int{ K.A^(3-p)} where N is the number of primitive triples a, b, c and a^p = b^p +c^p and a all values up to A. (K is about .155, 0.152929 for p =1 and 0.159155 for p = 2) For example, this gives for a up to 10,000, P = 1 N = 15292928 (actually 15198740) and for a up to 1000 p = 2 N = 318 ( actually 319) p > 2 N = 0 with any values for A and K apart from infinity. A better expression for N is N = int{ K.A^(2^(2-p))}. This is applying an RMS fit to the plot of N against A and A^p -B^p -C^p = 0
For p =2 K = 1/2.pi according to Alon Amit in Quora.
For p = 1, I get K about 0.152.
with A = 1,000,000 n = 151,981,776,195
For p = >2, N = 0 as long as K is less than one. I wonder whether there is a theoretical relationship between p and K?

Last edited by skipjack; March 29th, 2019 at 06:13 PM. March 29th, 2019, 03:13 PM #5 Member   Joined: Aug 2015 From: Chiddingfold, Surrey Posts: 57 Thanks: 3 Math Focus: Number theory, Applied maths Sorry I've got that wrong. Too old (91) and late at night. N = K.A^(3-p) is what I consider to be logical based on extensive examination of a = b + c and a^2 = b^c + b^c. Then for p = 3 we get N = K where N is the number of relatively primitive triples a, b and c. No idea what K is for p > 2 except perhaps always less than one. Tags fermat, theorem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post matqkks Number Theory 1 August 25th, 2018 06:38 AM Lourie Number Theory 3 April 1st, 2017 12:37 AM Rachanesamir Number Theory 2 May 13th, 2015 07:46 AM mente oscura Number Theory 10 June 6th, 2013 08:33 PM xfaisalx Number Theory 2 July 25th, 2010 04:59 AM

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