Re: Bring Math Arguments against this FERMAT LAST THEOREM PROOF

I will do this ,just as an exemple of how the Fermat last theorem proof must be understanded from previous presentstion of it.

Let's take X,Y, Z As being Integers pairwise coprime.
Let"s take X*Y*Z not divisible by 3
The equation X^3 +Y^3=Z^3 is impossible.

PROOF:
We write: X+Y-Z=B
Y-Z=-Q
Therefore: X-Q=B
X=Q+B
We write :X-Z=-P
Y-P=B
Therefore Y=B+P

Now we write:
(X+Y)*(X^2-X*Y+Y^2)=Z^3
Let's take :X+Y=W and
X^2-X*Y+Y^2=R
R can be rewwritten in another two ways:
R=(X+Y)^2-3*(X+Y)*Y+3*Y^2
and R= (X+Y)^2-3*(X+Y)^2+3*X^2
If (X+Y) is divisible by prime=m then (X+Y)=W and R have as common divisor prime=m if 3*Y^2 is divisible by prime=m and
3*X^2 is divisible by prime=m
That meanns Y^2 and X^2 must be divisible by prime=m
Therfore X and Y must be divisible by prime=m
That means that X and Y have as common divisor prime=m
That is impossible because X and Y are given as hving
no common divisor.
Therefore W=(X+Y) and R do not have any common divisor prime=m
Therfore W and R are relative prime.

But W*R=(X+Y)*(X^2-X*y+Y^2)=Z^3
Since W and R do not have any common divisor they are powers of 3
that is W=X+Y=u^3 and R=z^3 and Z=u*z
But W=X+Y=2*B+Q+P=u^3
and Z=B+Q+P=u*z
Therfore X+Y-Z=B=u^3-u*z
Therfore B=b*u
Therfore X+Y=2*b*u+Q+P=u^3
and Z=b*u+Q+P=u*z
Therefore Q+P =u*s and z=u^2-b

Now we write (B+Q)^3+(B+P)^3=(B+Q+P)^3=Z^3=(u^3)*(z^3)
(2*B+Q+P)*[(B+Q)^2-(B+Q)*(B+P)+(B+P)^2]=(u^3)*(z^3)
Since 2*B+Q+P=u^3 we have
z^3=[(B+Q)^2-(B+Q)*(B+P)+(B+P)^2]
If we expand the parantheses in above expression we get :
NR1. :z^3=B^2+2*(B)*(Q+P)+Q^2-Q*P+P^2

Let's write :
(-Z+P)^3+(-Z+Q)^3=-Z^3
(-2*Z+Q+P)*[(-Z+P)^2-(-Z+P)*(-Z+Q)+(-Z+Q)^2]=-Z^3=
=(-u^3)*z^3
We get that:
{(-Z+Q)^2-(-Z+Q)*(-Z+P)+(-Z+P)^2=z^3
OR NR2: (-Z)^2+ (-Z)*2*(Q+P)+Q^2-Q*P+P^2=z^3
We see that we could have got NR2. by Substituting B with (-Z) in NR1.
>From NR1 and NR2 we get that :
Z*B*h(Q'P)+Q^2-Q*P+P^2=z^3
We see that Q^2-Q*P+P^2=z*M
Let's multiply the above expression by (Q+P) =u*s:
(Q+P)*Z*B*h(Q,P)+(u*s)*z*M=u*s*z^3.
Now we divide the above equation by Z and get:
(Q+P)*B*h(Q,P)+ s*M=s*z^2
We know that s=z-b and z=u^2-b) and using them in the above
equation we get:
(Q+P)*B*h(Q,P)+s*M=(z-b)*z^2=z^3-b*z^2=((u^2-b)^3-b*z^2=
=(u^3)^2-3*b*(u^2)^2+3*(u^2)*(b^2)-b^3 - b*z^2=
= (2*B+Q+P)^2-3*b*(u^2)^2+3*(u^2)*b^2-b^3 - b*z^2
=Q^2+k*q*p+P^2
Therefore:
(Q+P)*B*h(Q,P)+s*M=Q^2+k*q*p+P^2
>From above Expression we get that s*M=Q^2+d*q*p+P^2
See OBSERVATION2 TO understand the last lines from above.
We know that u*z*s*M=Q^3+P^3=(Q+P)*(Q^2-Q*P+P^2) and now we got
a new expansion:
Q^3+P^3=Z*s*M=(Q+B+P)*(Q^2+d*q*p+P^2) where B is divisible by q and p
But according to Lame expansion in liniar factorsof
the polynom Q^3+P^3 that is impossible .
Therfore this proves FLT for n=3 when X*Y*Z is not divisible by3
The same way for any n=prime we get the FLT proof for X*Y*Z not
divisible byn
OBSERVATION2:
>From the Equations Z^3-X^3=Y^3 and
Z^3-Y^3=X^3 we get thatZ-X= P=p^3 and
Z-Y Q=p^3
and that B=b*u is divisible by q and p the same way we got
that X+Y=u^3 and that B is divisible by u.
Therefore q and p divide b.
I will show it if is not understood what I am saying.
This is written today August 30-05as an specific exemple
of Fermat Last Theorem proof which has been presented
as ideea .path and final argument on august 06 and
august 23.
created by GHEORGHE GHIATA
.

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