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March 23rd, 2019, 04:19 AM   #1
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Smile 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.
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March 23rd, 2019, 08:09 AM   #2
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Quote:
Originally Posted by magicterry View Post
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.
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March 23rd, 2019, 01:34 PM   #3
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Originally Posted by topsquark View Post
What are N, K, and A supposed to be?
-Dan
Thou shalt find them 3 critters in the alphabet...
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March 28th, 2019, 03:43 AM   #4
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Quote:
Originally Posted by magicterry View Post
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.
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March 29th, 2019, 03:13 PM   #5
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Angry

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.
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