Re: strategy for an instant, elementary proof of FLT


Cracked it - A necessary condition for your identity is p <= 3 :

john_rams...@xxxxxxxxxxxxxx wrote:
>
> quasi wrote:
> >
> > [...]
> >
> > f^p+g^p+h^p=k*(x^p+y^p+z^p) where k(x,y,z) is a polynomial
> > with integer coefficients which is not a perfect p'th power.
>
> [...]
>
> My approach assumes only that f, g, h are relatively prime
> as polynomials, although not necessarily coprime (relatively
> prime in pairs). The nature of k (whether or not it is a p-th
> power for example) isn't involved.
>
> For odd p, given your supposed identity, in the plane x + y = 0
> the surface f^p + g^p + h^p = 0 [1] has a p-fold root at z = 0.

[corrected "curve" to "surface" in the above line]

> Thus, as p > 1 we have f^(p-1).@f/@z + ... = 0 [2] at z = 0
> (where "@" denotes a Greek alpha).

Hereafter f, @f/@z, .., evaluated at z = 0 are polynomials
in one variable, say x, to which the "deg" function refers,
and deg(@f/@z){z=0} _equals_ deg(f){z=0}.

[1] and [2], considered as a pair of linear homogenous
equations in f^(p-1), g^(p-1), h^(p-1) then imply for
some pair of relatively prime polynomials u, v:

either:

u.f, u.g, u.h = v.@f/@z, v.@g/@z, v.@h/@z

or:

u.f^(p-1) = v.(g.@h/@z - h.@g/@z)

u.g^(p-1) = v.(h.@f/@z - f.@h/@z)

u.h^(p-1) = v.(f.@g/@z - g.@f/@z)

In either case, because f, g, h are relatively prime, we can
assume that v is constant.

Denoting d_f := deg(f), the first case requires d_u + d_f = d_f,
so that u is constant and consequently f must be a multiple of
e^(v.z/u), which is impossible.

For the second case, we can assume WLG that d_f <= d_g <= d_h,
so that d_f + d_g - 1 < 2.d_h <= (p - 1).d_h <= d_u + (p - 1).d_h

But by the last, d_u + (p - 1).d_h <= d_f + d_g, with equality
only if no cancellation of leading terms occurs in the RHS.
Thus necessary conditions are p <= 3 and u constant.



Cheers

John R Ramsden

.

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