# Fermat's last theorem

#### magicterry

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 a moderator:

#### topsquark

Math Team
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 a moderator:

#### Denis

Math Team
What are N, K, and A supposed to be?
-Dan
Thou shalt find them 3 critters in the alphabet...

#### 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 a moderator:

#### magicterry

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.