fermat'slasttheorem

Fermat's Last Theorem

a**n+b**n=c**n

=> n<3 [abc<>0]

Proof:

Basis for induction always exists (for any p):

a**4+b**4==c2**4 [mod 4]

Choose any even a, odd b, c2.

Inductive hypothesis:

a**(2**i)+b**(2**i)==ci**(2**i) [mod 2**i] (1)

=> there exists c(i+1) s.t. a**(2**i)+b**(2**i)==c(i+1)**(2**i) [mod 2**i] (2)

=> a**(2**i)+b**(2**i)==c(i+1)**(p2**i) [mod 2**i] [By Wanless' Theorem] [c(i+1)**(p2**i)==ci**(2**i) from (1)]

=> a**(p2**i)+b**(p2**i)==c(i+1)**(p2**i) [mod 2**i] [By Wanless' Theorem]

=> a**(2**(i+1))+b**(2**(i+1))==c(i+1)**(2**(i+1)) [mod 2**(i+1)] [By Wanless' Lemma from (2)] [i>1]

By mathematical induction:

a**(p2**m)+b**(p2**m)==cm**(p2**m) [mod 2**m] [m>1] with: c(m+1)**(p2**m)==cm**(2**m) [mod 2**m]

So:

Am**p+Bm**p==Cm**p [mod 2**m] [m>1] with: C(m+1)**p==c(m+1)**(p2**(m+1))==cm**(2**(m+1))==Cm**2 [mod 2**m]

Let m->99999…..:

Am**p+Bm**p=Cm**p with: C(m+1)<Cm [if p>2]

So, there is no smallest Cm, and therefore, by Fermat's Method of Infinite Descent, no non-zero A, B, C.

But, no pth power [p>2] (together with Fermat's work for n=4) => no nth power [n>2].