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From: smallfrey on 17 Jul 2010 13:07
Probabilistic arguments and empirical evidence indicate the probability for a given p, 6 divides p+1, that there do not exist a, b, c, such that a^p+b^p=c^p(mod p^2), p does not divide abc, is about .85. (See chapter 10 of Paulo Ribenboim's "Fermat's Last Theorem For Amateurs". A simpler argument than Ribenboim's shows that the probability is e^(-1/6). The "6" comes from the cross-ratio function.)

Then there's Legendre's version of Sophie Germain's theorem. Let p,q be distinct odd primes and assume that the following conditions are satisfied: (1) If a,b,c are integers such that a^p+b^p+c^p=0(mod q) then q divides abc. (2) p is not congruent modulo q to the pth power of an integer. Then the first case of Fermat's theorem is true for the exponent p.

The gist of the above is that a little progress on FLT can be made using elementary methods. The main topic here is Abel's conjecture. In 1823, Abel stated that if x,y,z are nonzero relatively prime integers such that 0<x<y<z and x^n+y^n=z^n (n>2), then none of x,y,z are prime-powers. In 1946, Inkeri showed that if 0<x<y<z, if p is a prime number, p does not divide xyz, and x^p+y^p=z^p, then z-y>1 (so that x cannot be a prime power). The following is an improvement (which might be new) of Inkeri's theorem.

If n>2, [(x^n + y^n)^(1/n) - x]^(1/n) + [(x^n + y^n)^(1/n) - y]^(1/n) > (x+y)^(1/n) where n is a natural number and x and y are positive real numbers. Barlow's formulae implied by a first-case solution of Fermat's equation a^p+b^p=c^p are (c^p-b^p)/(c-b)=R^p, (c^p-a^p)/(c-a)=S^p, (a^p+b^p)/(a+b)=T^p, c-b=r^p, c-a=s^p, and a+b=t^p where rR=a, sS=b, tT=c, and gcd(r,R)=gcd(s,S)=gcd(t,T)=1. (The prime factors of R, S, and T are of the form pk+1.) Then r divides t^p-s^p, s divides t^p-r^p, and t divides r^p+s^p. Then by the above inequality r+s>t>r,s and hence each of r, s, and t has a prime factor of the form pk+1. (Since t divides (r+s)*[(r^p+s^p)/(r+s)] and t>r+s, t and (r^p+s^p)/(r+s) must have a common factor.) Each of a, b, and c must then have at least two prime-power factors.

In 1909, Wieferich proved (using non-elementary methods) that a first-case solution of Fermat's equation implies 2^(p-1)=1(mod p^2). (The only known p that satisfy this condition are 1093 and 3511.) One of a,b,c must be even; assume that it is c. By the above improvement of Inkeri's theorem and the binomial theorem, a^p=1(mod p^2) if a has fewer than three prime-power factors and b^p=1(mod p^2) if b has fewer than three prime-power factors. Similarly, if 2^k divides c, 2^(k+1) does not divide c, and c has fewer than three prime-power factors other than 2, then c^p=2^(kp)(mod p^2). Then 2^(kp-1)=1(mod p^2). Under these restrictions, elementary methods can be used to arrive at Wieferich's criterion.
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