Re: Bring Math Arguments against this FERMAT LAST THEOREM PROOF

On 28 Aug 2005 05:41:14 -0700, matt271829-news@xxxxxxxxxxx wrote:

>quasi wrote:
>> On 25 Aug 2005 14:52:20 -0700, matt271829-news@xxxxxxxxxxx wrote:
>>
>> >
>> >george ghiata wrote:
>> >> > OBSERVATION:
>> >> > X,Y ,Z relative Prime numbers
>> >> > Let's say that X^3+Y^3=Z^3
>> >> > Let"s take:
>> >> > X+Y=W
>> >> > and
>> >> > R=X^2-X*Y+Y2=W^2-3*W*Y+3*Y^2=W^2-3*W*X+3*X^2
>> >
>> >You mean R=X^2-X*Y+Y^2, but yeah, OK.
>> >
>> >>
>> >> > When Z is not divisible by 3 we see that (X+Y)=W
>> >> > and R do not have any common divisor .
>> >
>
>[snip verbiage]
>
>>
>> Here is a proof of that part:
>>
>> Assume:
>>
>> (1) X,Y,Z are relatively prime
>> (2) X^3 + Y^3 = Z^3
>> (3) Z is not a multiple of 3
>>
>> Then R,W are relatively prime, where R=X^2-X*Y+Y^2 and W=X+Y.
>>
>> proof:
>>
>> Suppose instead that (R,W)>1. Let p be a common prime factor of R,W.
>>
>> p|W and W*R=Z^3 => p|Z => p .ne. 3 [since Z is not a multiple of 3]
>>
>> d|W^2-R => d|3*X*Y => p|X*Y [since p .ne. 3] => p|X or p|Y.
>
>I assume you mean p|W^2-R => p|3*X*Y => p|X*Y
>
>>
>> Without loss of generality, assume p|X. But then since also p|Z, the
>> equation Y^3=Z^3-X^3 implies p|Y^3, hence p|Y, contradicting the
>> assumption that X,Y,Z are relatively prime.
>>
>
>Great!
>
>Then
>
>(R,W) = 1 and R*W = Z^3 => R and W are perfect cubes.
>
>So R = z^3 and W = u^3 for some integers z and u.
>
>And
>
>Z^3 = R*W = z^3*u^3
>=> Z = u*z
>
>Continuing with GG's proposed proof:
>
>> When Z is divisible by ,that is Z=u*3*z we see that
>> W and R have as common
>> divisor only 3
>> Therfore W=(u^3)*3^(3-1) and R=3*z^n
>
>I assume this means "When Z is divisible by 3..."
>
>Anyone care to offer a proof of this next bit?

Assume 3|Z.

Then X,Y are not multiples of 3 since if 3 divides one of them, the
equation X^3+Y^3=Z^3 together with 3|Z would imply 3 divides the
other, contrary to the assumption that X,Y,Z are relatively prime.

Hence X*Y is not a multiple of 3.

Now W*R=Z^3 => 3|(W*R) => 3|W or 3|R.

Then the relation R=W^2-3*X*Y => R=W^2 mod 3 => both R,W are multiples
of 3 (since we know at least one is).

But 3|W => 9|W^2 => 9 does not divide R (since 9 does not divide
3*X*Y). Hence R is a multiple of 3 but not a multiple of 9.

Then W*R=Z^3 => W*R is a multiple of 27 so W must be a multiple of 9.

Since we know that in general (W,R)=1 or 3, in this case we must have
(W,R)=3. Then the integers W/9,R/3 are relatively prime and since the
product (W/9)*(R/3) is a perfect cube (=(Z/3)^3), it follows that R/3,
W/9 are both perfect cubes.

So one can write W=9*u^3, R=3*z^3 and then Z can be expressed as
3*u*z, as claimed.

So this part of the proof is also correct.

Without looking too carefully, it appears that in the next parts of
the proof we are about to be dragged through a lot of elementary
algebra -- substitutions, simplifications, etc, all of which is
probably correct. This partly camouflages any later flaw, since one
has to actually go through all the algebraic preliminaries to get at
whatever is the key proof idea.

quasi
.

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