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.

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.

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