Harristotelian Logic

There have been several recent claims of counter examples and/or counter arguments to the short proof of Fermat's Last Theorem.

I noted that those claims contradicted an independent verification of a key piece of my work, which I called Area One.

Some questioned that, and continued to make said claims.

Here's a refutation, which focuses on expressions that follow from

x^3 + y^3 = z^3

for simplicity.

It assumes some familiarity with the proof.

Consider

(v^3+1)z^6 - 3v x^2 y^2 z^2 - x^3 y^3 =

w1 w2 w3(s1 z^2 + t1 xy)(s2 z^2 + t2 xy)(s3 z^2 + t3 xy)

where s1 = s2 = sqrt(v+1), and

w1 s1 = a1, w2 s2 = a2, w3 s3 = a3, w1 t1 = b1,

w2 t2 = b2, w3 t3 = b3,

so

w1 w2 w3 (s1 z^2 + t1 xy)(s2 z^2 + t2 xy)(s3 z^2 + t3 xy) = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy),

where a1, a2, a3, b1, b2 and b3 are algebraic integers, and the ring is formed by adjoining the square roots of the integers, t1, t2 and t3 to the ring of integers.

Now there is also the factor f, which is a prime integer factor of sqrt(v+1) as described in the short FLT Proof.

(Reference http://www.msnusers.com/AmateurMath/fltproof.msnw )

Then the Area One result within the specified ring can be written as

(s1 z^2 + t1 xy)(s2 z^2 + t2 xy) = 0(mod f^{n+2j}),

which is

(sqrt(m)f^j z^2 + t1 xy)(sqrt(m)f^j z^2 + t2 xy) = 0(mod f^{n+2j}),

so

(sqrt(m) z^2 + t1 uy)(sqrt(m) z^2 + t2 uy) = 0(mod f^n),

This result has been independently corroborated.

I moved from a ring like the one here described to the ring of algebraic integers as I thought it gave a more beautiful proof, and was more in line with what I'd originally envisioned.

Within the ring of algebraic integers the Area One result is

(a1 z^2 + b1 uy)( a2 z^2 + b2 uy) = 0(mod f^n),

and posters have claimed that it is false.

But consider that

w1 w2 w3(s1 z^2 + t1 xy)(s2 z^2 + t2 xy)(s3 z^2 + t3 xy) = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy),

so their claim that a1 does not have a factor of f within the ring of algebraic integers requires that w1 s1 not have a factor of f either.

But, s1 = sqrt(v+1) = m f^j, so that position requires that f must be a unit in the ring I described above.

However, the correctness of the Area One result depends on f NOT being a unit in the ring, as if it is then

(sqrt(m) z^2 + t1 uy)(sqrt(m) z^2 + t2 uy) = 0(mod f^n),

is meaningless, and in fact, it has been independently shown to not be a unit in that ring.

In considering the arguments above the important point is to realize that the shift in rings does not change the *value* of the factors which is

(v^3+1)z^6 - 3v x^2 y^2 z^2 - x^3 y^3

so results in different rings must be linked.

What I think is of interest here is that f must be a unit in the ring for counter claims to be valid, which is a position that was taken, heatedly argued, and refuted. The refutation of the claim that f was a unit was accepted. Yet now a position that requires that f be a unit has been from what I've seen generally accepted.

In surveys on the newsgroup that I've conducted, posters seem to rely on the notion that there is a counterexample or counterarguments, which is still puzzling to me. I'm curious as to whether or not anyone can explain.

So, succinctly, certain posters made claims that required that f be a unit within the ring, those claims were refuted by someone other than myself, and the refutation accepted on the newsgroup. Then posters came back later, made arguments which required that f be a unit, as I've shown in this post, and these claims were accepted on the newsgroup.

However, though human beings can be inconsistent in this way, mathematics cannot, so their claims are false.