Recent content by rayb07

I've posted a different version here: http://groups.google.com/group/fermats-last-theorem-flt-2009

The idea is that if xyz is a multiple of p then y is a multiple of p. When the RHS of y^p = ((z - x) + x)^p - x^p is expanded, the first term can be divided by p(z - x) to leave an integer coprime to (z - x) while all remaining terms are multiples of p(z - x)^2, leading to the...

I found the flaw. Line [14] does not follow from [13]. If z is a multiple of p then p divides (x + y) (p -1 +pn) times, with n a positive integer. Both (xp)/(bp) and (yp/cp) have the form (pk + 1)p, so in the equation (x + y) = bp + cp + 2pabch, a is a multiple of p. Both z and x are...

In the RHS of zp = p(x + y)xp-1 - 1/2 p(p-1)(x + y)2xp-2 + ... + (x + y)p both p and x are coprime to (x + y), while the second term is a multiple of (x + y)2, the third term is a multiple of (x + y)3, and so on. Therefore dividing the first term by (x + y) leaves an integer coprime to (x +...

In what context? If zp can be divided by (x + y) to leave an integer coprime to (x + y), then (x + y) must be a perfect pth power. The proof makes assumptions (hopefully correct) about z/a and x/b. I see no reason why those assumptions should be correct for p = 2. Also, operations to make x...

zp = p(x + y)xp-1 - 1/2 p(p-1)(x + y)2xp-2 + ... + (x + y)p If z is coprime to p then the RHS can be divided by (x + y) to leave an integer coprime to (x + y), therefore (x + y) is a perfect pth power. I've corrected the proof to mention that the variables are all nonzero integers. The...

Below is the address of an attempted proof of FLT that I have not been able to find a flaw in. It works by increasing x, y, and z by a common integer factor so that, if x^p +y^p = z^p, with x + y = a^p and z - y = b^p, then (z/a) = (x/b)^((1 + 1/(p-1)) (mod g), with g a prime factor of y that...

Sorry, I don't have a proof, there's a simple error in it. My bad.

I think I have a short proof: http://rayb07.blogspot.com/2007/08/fermat-equation.html