The Faux Pas Proof of Fermat's Last Theorem

Date:	6/24/99 
Author: Greg@Tell-all.com (Gregory N.Mazur, www.tell-all.com)


Abstract:

For a single dimensional length, it is self-evident that a sum of equal
integers (x^n + x^n) cannot equal an odd integer, z or an odd z^n. This
special case of an odd integer z, raised to an integer power n is the basis
of a proof of the general case of Fermat's Last Theorem.

An even integer z raised to an integer power n, such as the diameter z^n,
may be bisected into two integer radii, x^n each. I prove that such even
integer diameters have prime facrors of z whereas their integer radii, as
x^n each have no prime factors of x. That is because x cannot be an integer
where such z^n = 2x^n.

Trearting the two radii as one integer I further prove, as Fermat said,
"...On the contrary, it is impossible to separate a cube into two cubes, a
fourth power into two fourth powers, or generally, any power above the
second into powers of the same degree."

1991 Mathematical Subject Classification: 11D41, 11D61

Key words and phrases: Fermat, Fermat's Last Theorem, prime factorization,
diameter, radii, Diophantine equations, Fundamental Theorem of Arithmetic,
dimension, powers, integers.

Article:

Fermat's Last Theorem states that there are no integers X, Y, Z and n, for
n > 2 such that a Diophantine solution in integers exists for X^n + Y^n =
Z^n. The following proof by contradiction shows that if Fermat's Last
Theorem were false, it would imply impossible contradictions.

Assume Z and n are possitive integers with n > 2.
Whether or not Fermat's Last Theorem is true, Z^n may be bisected into
equal halves.
First we will need to prove an intermediate result that  X cannot be an
integer when X = Y. 
For all Z^n a magnitude, R, exists such that R^n = (Z^n)/2.
Assume, until proven false, that R = an integer that permits a Diophantine
equation R^n + R^n = Z^n.
Let Z^n = a diameter of a circle.
Let R^n = a radius of that circle.
The Fundamental Theorem of Arithmetic requires that all integers have a
unique prime factorization. 
If Z is any composite number that has factors in common with R, then it is
possible to divide the left and right side of the equation by any factors
that they both share in common.
Assume the factors (p_1)*(p_2)* ... (p_i) = P, (P => 1) = the product of
all shared factors of Z and R.
Let (Z/P) = z.
Since Z is an integer, then z must be an integer.
Let (R/P) = r.
Since R is an integer, until proven false, then r must be an integer, until
proven false.
z^n = r^n + r^n.
If z is an odd integer, then z^n is an odd integer.
Assume, until proven false, that r can be an integer that permits a
diophantine equation in the form:
z^n = r^n + r^n.
If z^n is an odd integer, then z^n cannot equal 2r^n which is an even
integer.
Since z must be odd by definition in this case, no integer solution exists
for r or radius r^n when z is odd.

If z and z^n are even integers, then z must equal 2a, where a is an integer
=> 1.
(2a)^n = r^n + r^n = 2r^n.
(2^n)(a^n) =  r^n + r^n = 2r^n.
If z is even, we could divide and factor out a common prime, 2, from both
sides of the equation.
But this is impossible.
If r has no factors in common with z, then (r^n + r^n) cannot be factored
down any further.
If r and z have no common factor, it is impossible to divide the left and
right side of the equation by 2.

But z must equal equal 2a, an even integer, by definition in this case.

So r cannot be either an even integer or an odd integer.
If r cannot be an integer in z^n = 2r^n, then R has no prime factors.
Since P => 1, then r = R and z = Z where R cannot be an integer.
So there is no integer R that satsfies any diophantine equation in the form
Z^n = R^n + R^n.

This proves that Fermat's Last Theorem is true in the restricted case where
diameter Z^n  = R^n + R^n.

Now assume that one radius R^n may be stretched to the same extent that the
other radius is shortened.
Assume that the integers Z^n and n are held constant as the non-integer R
component varies.
Let d = an arbitrary number that simultaneously increases one radius while
decreasing the other.
Let X^n = (R + d)^n.
Let Y^n = (R - d)^n.
Let Z^n =  (R + d)^n + (R - d)^n.
If both (R + d) and (R - d) are integers, then both X and Y are integers,
which would prove that
Fermat's Last Theorem is false.
If Z^n is odd, then {X,Y} = {even,odd} or {odd, even}. So  (X + Y) = 2R =
an odd integer where R is one half of an odd integer; R = (X + Y)/2. 
Let A = 2R = X + Y = an odd integer. Then R = A/2.
R^n = (A/2)^n
2R^n = 2(A/2)^n = Z^n = an odd integer.
(A/2)^n = ((1/2)^n)*(A^n) = one half of an odd integer.
Since A remains the odd integer (X + Y), then A^n is an odd integer.
But (1/2)^n multiplied by an odd A^n yields a fraction whose numerator must
be an odd integer and whose denominator must be an even integer > 2 for all
n > 2.

But the denominator cannot be > 2 if the required (A/2)^n must be precisely
one half of an odd integer.
So R^n = (A/2)^n cannot be one half of an odd integer.

Therefore no such d variable exists that simultaneously converts both (R +
d) and (R - d) into whole integers at the same time where the defined odd
integer diameter  Z^n = 2R^n.

If Z^n were even, then X + Y = 2R = two even integers or two odd integers.
Both sides could be factored by 2 to yield a whole integer, R. This is
impossible because we proved that R cannot be an integer when Z^n = 2R^n.

In conclusion, no integer solutions exist for Z^n = X^n + Y^n when X, Y, Z
and n must all be integers with n greater than 2.

Q.E.D.




Gregory N. Mazur
Greg@tell-all.com