A Tangential Model of Fermat AMS

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Tangential Model of Fermat’s Conjecture

Tony Thomas

Abstract

This paper uses a geometrical construction to model Fermat’s Conjecture (FLT). The contrary

assumption ( FLT) is made and propositions derived from this assumption with the aim of

revealing inconsistencies under the conditions applicable to FLT.

Introduction

The aim of this paper is to establish a more direct proof of Fermat’s Last Theorem than the proof

published by Andrew Wiles in The Annals of Mathematics 142 (1995) . The main idea is that the

three terms a n , b n and c n all lie on the curve x n which facilitates the construction of diagrams

showing the relation between these terms and several auxiliary variables used in the proof. The

two diagrams used are shown in the annex.

d4

Fermat’s Conjecture

Fermat’s Conjecture: there are no natural numbers a, b, c, n such that a n + b n = c n when n > 2 .

FLT may be expressed formally as: abcn ℕ a n + b n = c n

subject to the following conditions:

(1)

a,b,c and n are distinct natural numbers* abcn ∈ ℕ

(2)

a,b and c have no common factors abc NCF

(3)

n is greater than 2. n>2

(4) a useful convention a<b<c

(5) n is a prime number

* The term integer is used throughout to signify a positive integer ie n ∈ ℕ

Parity limitations

Condition (2) implies that a , b and c cannot all be even numbers otherwise they would have the

common factor 2 . Furthermore, if any two variables are even numbers then they have the common

factor 2 and the third variable must also be even. Consequently, two of the variables must be odd

and the third even. It will be shown later that b must be an odd variable given condition (3) so

either a or c is even.



The geometrical model

Diagram 1 shows the main elements of the model. Given any two integers a and c there will be two

points (a, a n ) and (c, c n ) on the curve x n . A line drawn through these points intersects the

x-axis at γ . Parallel to this secant there is a unique tangent that meets the curve at (K, K n ) and

intersects the x-axis at α .

Let the gradient of this tangent be g such that:

− = g ,

− = g and

− = g

Furthermore it can be seen that−

− = g

Henceforth the antithesis of FLT will be assumed to be true: abcn ℕ a n + b n = c n subject to the

given conditions (1) through (5).

The Gradient as basis for the model

The gradient of the secant must be equal to the gradient of the tangent, since the two lines are

parallel.

g =

=

Δ

Δ consequently nK n-1 =

−

−.

Applying the antithesis FLT, b n = c n – a n produce the equation nK n-1 =

− .

It should be noted that−

− is always an odd integer when n is a prime greater than 2 and c and a

are of opposite parities; therefore g is an odd integer so b must also be an odd integer. Since n is

defined as a prime number greater than 2 it must be odd, so g must also be odd.

Superficially, it may seem from the equation b n = (c – a)nK n-1 that b must be irrational, however by

setting K = n(c – a) the integer solution b = K results as a special case.

It can be seen from Diagram 2 that g =

−β so that

−β=

− , consequently β = a + b – c where β

is the x-axis intercept of a line drawn parallel to the tangent line through the point (b, b n ) .

The value of

From Diagram 1 it is evident that

−α= g = nK n-1 so that n =

−α.

Rearranging; K =α

−1and α =

(−1)

. If K is an odd integer then α must be even, although α will

be a rational fraction if n | K . Also, n – 1 | α is necessary for K ∈ ℕ .

Henceforward it will be assumed for convenience that a is an even number and that c is odd until

reconsidered at the appropriate time.



Why has n been defined as a prime?

If FLT can be proven for prime exponents (n>2) then it must be true for all exponents (n 4) .

Suppose m ∈ ℕ and m = pq where p is prime and q is either prime or composite.

a m + b m = c m a pq + b pq + c pq

Let A = a q , B = b q and C = c q so that a m + b m = c m A p + B p = C p

If it can be shown that ABC ℕ A p + B p = C p then it follows that abc ℕ a m + b m = c m

Consequently, it is only necessary to prove FLT for prime exponents.

The value of

Lemma: g | a n g | c n

g | b n (g | a n g | c n )

Condition (2) stipulates a , b and c have no common factors.

Since b n = g(c - a) and g ∈ ℕ it follows that g | b n .

Applying modus ponens proves the Lemma

From Diagram 1 it can be seen that a – γ =

and c – γ =

So γ = a –

and γ = c –

Applying the Lemma shows γ is a rational fraction.

However, g γ = ga – a n and g γ = gc – c n so it is evident that g γ is an even number.

Analysis of

When n= 2 , K lies at the-mid-point between a and c so K =+

2 because

−

−=

(+)(−)

−

therefore c + a = 2K = g.

This special characteristic explains why Pythagorean triples are possible because a and c always

lie symmetrically about K . It will be shown below that this is not the case when n >2 so the secant

can never simultaneously intersect integer values of (a, a n ) and (c, c n ) when K ℕ .

The variables a and c can be defined in terms of K as follows:

a = K – q and c = K + p so that c – a = p + q , where p and q are deviations on either side of K .

A secant may be formed by moving the tangent an integer distance left of K .

When n =2 an equal deviation occurs to the right of K so that p = q . The underlying reason is that

the differential coefficient of x 2 is the rectilinear function 2x . When n > 2 it is evident that an

integer move to the left will result in a lesser and possibly fractional deviation to the right.



It can be seen from Diagram 2 that g =−

− =

−

−. In terms of the deviations this can be

expressed as: p =(−)

−and q =

(−)

− , where Z = K n + g( α – γ) = g(K – γ) .

Parities of variables

Case 1 Case 2 Functiona even odd

c odd even

g odd odd nK n-1

Z odd odd g(K – γ)

p even odd c – K

q odd even K – a

g α even even (n-1)K n

gγ even even ga – a n

K odd oddα

−1

n odd odd

γ fraction fraction a –

Parity analysis of and q

Case 1: p =(−)

− c – K =

(−)(−)

−

(o-o) =

(−)(−)

− e =

The result is a disparity. The variables on the RHS are assumed to be integers so the conclusion is

that p cannot be an integer.

Case 2: q =(−)

− K – a =

(−)(−)

−

(o-o) =

(−)(−)

− e =

This is also a disparity showing that q is not an integer assuming the variables on the right are

integers.

Case 1 assigned an even parity to a and an odd parity to c . Case 2 assigned an odd parity to a and

an even parity to c . This reversal had no effect on the results.

Corollary: Either p is not an integer or q is not an integer under the given assumptions.

So (p + q) ℕ and since p + q = c – a it follows that (c – a) ℕ .



This conclusion is a consequence of assuming that K is an integer and also that Z , as a function of K

is an integer. The remaining possibilities are that K is a rational fraction or that it is an irrational

number. These possibilities will now be examined.

Assuming is a rational fraction

Let K =

where u, v ℕ and u, v have no common factors.

g = nK n-1 = (

)

Since g, n ℕ and

ℕ g cannot be an integer as v n-1 | n because n is prime.

Corollary: If K is a rational fraction then g cannot be an integer, but g must be an integer if

c, a ℕ because−

− ℕ and g =

−

− .

If g is not an integer then either a is not an integer or c is not an integer.

Assuming K is an irrational number

The above proof will not suffice when K ℚ because K may be an n-1th root. An auxiliary analysis

of K is required first.

Let K ℚ and let K = n where ℚ

Now n =

−αso n =

− (−1) and the irrational term cancels out.

This resolves the rationality of n when K is irrational.

The above method can now be applied when K ℚ : g = nK n-1 = n(n ) n-1 = n n n-1

but as noted above g = nK n-1 =

−α.

Substituting n for K : g = n(n ) n-1 =( )

− (−1)=

( )

=n n n-1 .

So g will be irrational unless n-1 ℚ

Let K be the n-1th root of m , where m ℕ

So K = n = m 1/n-1 and n n-1 n-1 = m

Therefore n-1 =

which is a rational fraction.

Now α = (n-1) so n-1 =

=

(−1)

therefore m =

(−1) which is consistent with K =

−1 = n

but α n-1 = n-1 (n-1) n-1 = (n-1) n-1

which is a rational fraction.

No inconsistency arises when K ℚ in the special case when K = n = m 1/n-1 .



This is intuitively evident since a secant passing through the integer points (a,a n ) and (c,c n ) must

be associated with a tangent which can be irrational without affecting the rationality of (c – a) .

Introducing Fermat’s Conjecture

−α= g and m = K n-1 so

− = g

but n =

−αso

− = mn consequently mn = g

but g = −

− =

− given FLT so b n = mn(c-a) and b n = K n-1 n(c-a).

Now n cannot have a rational nth root because it is a prime and the nth root of K n-1 is also

irrational.

− ℕ provided that c n – a n = b n but b = (c – a) 1/n (mn) 1/n is irrational whether or not (c - a) 1/n is

rational. Given that the secant passes through the integer points (a, a n ) and (c, c n ) when K ℚ

the conclusion must be that (c – a ) ℕ and that b ℚ and consequently that c n – a n b n .

Summary and conclusion

(1) If K is an integer then g is not an integer so either a is not an integer or c is not an integer.

(2) If K is a rational fraction then g is not an integer so either a is not an integer or c is not an

integer

(3) If K is an irrational number then b is irrational.

In arriving at these conclusions the negative hypothesis abcn ℕ a n + b n = c n was assumed

together with the conditions (1) through (5) applicable to FLT. The three conclusions above prove

that the negative hypothesis under the given conditions is inconsistent, consequently FLT

must be true reductio ad absurdum .

Tony Thomas

7 Sunblest Crt

Eatons Hill

Queensland 4037

15th November 2015



Annex

Documents

A Tangential Model of Fermat AMS