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^(3p)} 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  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,266 Thanks: 934 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan Last edited by skipjack; March 29th, 2019 at 06:14 PM.  
March 23rd, 2019, 01:34 PM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  
March 28th, 2019, 03:43 AM  #4  
Member Joined: Aug 2015 From: Chiddingfold, Surrey Posts: 57 Thanks: 3 Math Focus: Number theory, Applied maths  Quote:
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^(3p) 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. 

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