Monday, February 09, 1998


New proof of FLT (sigh)

Made an earlier hasty post on sci.math but luckily with the delay of these posts it's not even out there yet. This looks similar but I realized that I had stumbled upon something useful while previously it appeared I'd only made another mistaken post. Damn signs.


Given x^p + y^p + z^p = 0; x,y,z relatively prime, p odd prime, no solution exists;

flt (Fermat's Little Theorem)

x^p = x(mod p), x E Integer; p prime;

(often used below but not specifically cited)

General Proof of FLT

(a+b+c)^p = a^p + b^p + c^p + p(a+b)(a+c)(b+c)Q; a,b,c, Q E Int; p odd prime

(stated without proof because I've proven it enough over the years and I'm tired)

let j=(f-1) where f is a prime factor of z not contained in (x+y)

let a = x^j, b = y^j, c=z^j

(i) (ii) (iii)
(x^j + y^j + z^j)^p = x^{pj) + y^{pj) + z^{pj} + p(x^j + y^j)(x^j +z^j)(y^j + z^j)Q

Consider inside the term on the left (i)

x^j + y^j + z^j = 1+1+0 (mod f)

Consider (ii)

x^{pj) + y^{pj) + z^{pj} = 1+1 +0 (mod f)

Therefore, what's left on the end (iii) must be divisible by f

p(x^j + y^j)(x^j +z^j)(y^j + z^j)Q (again note that x^j+ y^j = 2(mod f) because j is even)

Which means that there's nothing left to be divisible by f but Q and there's not even that for p=3.

But, of course, since x^p + y^p + z^p = 0; (x+y+z)^p = p(x+y)(x+z)(y+z)Q and relative primeness is still broken.

Almost missed it but it doesn't really matter; I guess. Damn signs.

Oh yeah, I remember noticing years ago that it was easy to prove that z couldn't be prime. Felt it had to come in handy somewhere. And, I even had an intuition that evenness had something to do with this as well.

Pretty nifty. Don't bother sending correspondence— I won't read it. I handle the delete key very well. If I finally managed to be right this time, I figure I'll eventually find out, one way or another.

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