IN DEFENSE OF MR. FERMAT

During the course of studies on the Goldbach Conjecture, using finite methods, what seems to be an elementary proof of Fermat’s “Last” Theorem has been found. Astonishing here is the lucidity of the arguments and immediacy of their logic. Hopefully, by (numeric) application to the so-called “hard” problems of Number Theory, some manner of agreement (disputation) will arise.

FERMAT’S LAST THEOREM

Suppose the following equation has the solution for positive r, a and b.

(1) r^n = a^n + b^n where a>0 and b>0.

Clearly the following congruences must hold (implied).

(2) r^n - b^n = 0 (mod a^n)

(2)’ r^n – a^n = 0 (mod b^n)

From now on forsaking implicit-only relations (2) and (2)’ can be used to further specify the consequences of (1), using the single valued function,F(x,y)

F(x,y)

Let {c} represent, i.) c.GE. 0, the greatest Integer in c, or ii.) c< 0,. -({|c|} + 1).

F(x,y) is Defined y: y.NE.0, F(x,y) = x – ( |y| * { x / |y| }), y: y=0 F(x,y) is undefined. F may be referred to as the “least positive (or 0) remainder of x on division by |y| function.”

Explicitly now as [(1) implies (2)] and [(1) implies (2)’] when [r,a,and b] are positive, (3) F((r^n – b^n), a^n) = 0 as well as (3)’ F((r^n,a^n), b^n) = 0.

ORDER

The following “order” relations are definitive of the assumption [a > b] when [r, a and b] are positive.

(4) F(r^n,a^n) – b^n = 0

(4)’ F(r^n,b^n) – a^n .NE. 0

PROOF

Let [r,a,and b] be positive.

First suppose a^n = b^n.Then, (1) can be represented r^n = 2*(a^n) which has no solution in positive in positive Integers

Let [a^n > b^n]. Then F(r^n,a^n) = b^n and (4) immediately follows. Regarding (4)’,

F(r^n,b^n) = F(a^n,b^n) .NE. a^n so that:

F(r^n,b^n) – a^n = F(r^n,b^n) – F(a^n,b^n) – d*(b^n) .NE. 0

where d .GE.1.

Conversely, suppose [a^n is not greater than b^n] while (4) holds. Then [b^n > a^n] and transformation of a to b and b to a has no effect on (1) but “rectifies” the representations of (4) and (4)’ accordingly as [r>b>a>0]. Thus, (4)’ rightfully becomes, F(r^n,a^n) – b^n .NE. 0.! which is inconsistent with prior (4). Proof of the case (4)’ follows complementarily. ! denotes contradiction. Thus, a^n > b^n may be defined in this manner.

Prominent now becomes the anticipation that (1) has a distinct ordering of a^n and b^n is contradictory. A-priori however must be the proof that if (1) is solvable for some n such that n is greater than or equal to 3, then each case of n may be “rewritten” (in possibly different r, a, b and n) as v^n’ = u^n’ + w^n’, where n’ is some divisor of n such that [v>u>w>0] is necessary. The different cases of n will eventually be considered n’ such that n’ is greater than or equal to 3.

OBJECT: [(1) holds for any n such that n is greater than or equal to 3] implies (1) may be rewritten (in possibly different r, a, b and n) as v^n’ = u^n’ + w^n’, where n’ is greater than or equal to 3 such that [v, u and w .GT. 0 ] is necessary.

CASE 1: n has an odd divisor, q, greater than 1.

Let (1) be written, r^[(n/q)*q] = a^[(n/q)*q] + b^[(n/q)*q] and substituted as v^q = u^q + w^q. If n/q is even, the fact that plus or minus r, plus or minus a and plus or minus b are roots of [v, u, w] implies the latter are all positive. On the other hand, if n/q is odd, original conditions, [a>0 and b>0] imply the relation in [v,u,w,n’] has strictly positive exponential bases.

CASE 2: n = 2^t where t is greater than or equal to 2.

Let(1) be written;

[r^2^(t-1)]^ 2 = [a^2^(t-1)]^2 + [b^2^(t-1)]^2 which upon substitution becomes v^2 = u^2 + w^2. As before, the fact that 2^(t-1) is even and the existence of roots plus or minus r, plus or minus a and plus or minus b, corresponding to [v,u,w] imply they are all positive. Note that the case n=2 cannot be rewritten such that its root is not possibly negative.

Having limited the existence of a positive-only solution for a rewrite of (1) such that [r,a,b,n] goes to [v,u,w,n’] for all n greater than or equal to 3, order is specific and consequently subject to contradiction.

Suppose now that (1) has been rewritten for some n, where n is equal to or greater than n’ so that [r,a,b and n] is transformed to a reduced form, redefining [r,a, b and n] to be [u,v,w and n’]. Thus n’ is greater than or equal to 3 implies [v>u>w>0]. Consider the transformation (under n’), T, such that T:v to v, u to w, and w to u. Respecting (1), T(v,u,w) = Tnot(v,u,w) where Tnot signifies not T. It follows that [u>w] can be transformed to [w>u] such that [u>w] is unaltered.! This can happen only for the case,u=w, which is moot. Conclusively, order must be assignable when it exists. ( ! denotes contradiction.).

SUPPOSE: [r is conjugated negative in sign to [a]] which implies [(4) and (4)’ are indefinite] allows a solution to (1) for the case n=2.

EXAMPLE

Let r^2 = a^2 + b^2 (n=2). Divide through by q^2, where q^2 is the greatest (square) divisor common to [r^2,a^2 and b^2] to obtain the so-called “primitive” form, r^2/q ^2 = a^2/q ^2 + b^2/q ^2. Rearrange as follows, conjugating plus or minus |r| with plus or minus |a| where they are of different sign respectively:

|a|^2/q ^2 * [(-|r|)^2/|a|^2 – 1] = (|a|^2)/q ^2)*[ |r|^2/a^2 – 1] = b^2/q ^2

b^2/q ^2 is substituted by 0, such that a^2/q ^2, as an Integer multiple of 0, drops out. F is applied such that the remaining is a relation among squares (preserving n). For example,

F[(-5)^2/(4)^2 , 9] = F[4^2/4^2 , 9] = F[5^2/(-4)^2 , 9] =

F[(-4)^2/(-4)^2 , 9]

Consequently, F[1^2 , 9] – (-1)^2 = 0 is identifiable as a relation among squares. Thus the presence of values altering the positivity of [r,a and b] allows a solution to (1).

In the event that [v>u>w>0] (under n’) the latter method clearly cannot succeed. In any case, the instances such that (1) is rewritten (for all non-trivial n, 3 or greater) stand consistent while example accordingly excepts the case, n=2.

KEY

! denotes contradiction

|x| denotes the absolute value of x

* denotes multiplication

{x} denotes the greatest Integer in x when x = |x|

^ denotes exponentiation

Tnot denotes “not T”

.LE. denotes “less than or equal to”

.GE. denotes “greater than or equal to”

.NE. denotes “not equal to”