### 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.

FLT

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

then

(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.

FLT

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

then

(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.