
Re: FLT Discussion: Simplifying
Posted:
Jan 19, 2001 10:11 AM


In article <94810q$6o1$1@nnrp1.deja.com> jstevh@mydeja.com writes: ... > Some have tried to find fault with my method using the n=2 case, which > *does* use x+y+vz. I think those are interesting, so I'll go over them > quickly.
Interesting that you find them interesting now, but not in a reply to the posting where I wrote about that example.
> x+y+vz = 0(mod (x+y+vz)), jumping ahead > x^2 + 2xy + y^2 = v^2z^2, and jumping to the end, > (v^2  1)z^2  2xy = 0(mod (x+y+vz)), with x^2 + y^2 = z^2. > Some have argued, why not factor what's on the left to get > (sqrt(v^2  1)z  sqrt(2xy))(sqrt(v^2  1)z + sqrt(2xy)) = 0, > so (sqrt(v^2  1)z  sqrt(2xy)) = 0(mod (x+y+vz)?
But this is a bit of a distortion, as I came with: [(1v^2)z  sqrt((1v^2).2xy][(1v^2)z + sqrt((1v^2).2xy)] = 0. > > Is there a really simple answer that's rigorous? I guess so, but the > answer I give is you look at the exponents.
Yup, I do and see that there are no integral exponents for x, y and v. So what? In your factorisation v does not have an integral exponent. (Yes, I know that you say that sqrt(x^2 + y^2) can be written as x + ... + y, but you do not bother to fill in the ellipses, and the only thing I can fill in there sensibly is [sqrt(x^2+y^2)xy]. Like, sqrt(x^2 + y^2) is a polynomial, which is *false*.) > > To me it's a simple matter of asking what expression can multiply times > (x+y+vz) to get (sqrt(v^2  1)z  sqrt(2xy))?
Yup, so what is that expression?
> Again, the statements are true if x^p + y^p = z^p. Note, as written > that is saying that that statement is true. That is, that we have some > x, y and z that fit.
No. Not all statements are true in your proof. In particular one statement is *not* true, but false. As I parafrased it in my "proof" (replace 'n' by '1' to get your proof, and note that this is a "valid" statement for *all* values of 'n'): [(v^5+n)z^2  (5v^3+sqrt(5v^620nv)*xy/2)] * [(v^5+n)z^2  (5v^3sqrt(5v^620nv)*xy/2)] = 0 (mod (x+y+vz)/h), when x^5 + y^5 = nz^5. Fill in n=68101, x=15, y=17, z=2, h=2 (I once asked whether that was the only coprime solution for n=68101, and indeed, it has been proven) to get: [4v^5+272404  (5v^3+sqrt(5v^61362020v).255/2] * [4v^5+272404  (5v^3sqrt(5v^61362020v).255/2] = 0 (mod v+16). and note that even this is "valid". But none of the factors is divisible by (v+16) by any definition that I know. (Note that when you fill in v=16 on the left the product is indeed 0, proving that is is "valid".)
> http://www.mindspring.com/~jstev/FLTp5.htm.
I still await your response about which statement in the "proof" on http://www.cwi.nl/~dik/mathematics/jsh.html is exactly wrong, and why.  dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

