Fermat’s Last Theorem Revisited

Fermat’s Last Theorem Revisited

Introduction

Andrew Wiles published a proof of Fermat’s Last Theorem in The Annals of Mathematics, 142 (1995). This long work drew upon several arcane branches of mathematics, most of which are accessible only to professional mathematicians. The hopes for a simpler proof that could be attributed to Fermat himself remain unsatisfied. As a consequence of legions of failed proofs, the use of Pythagorean triples in the solution has been denigrated as the mark of amateurs, despite the fact that Fermat was engaged in studying them when he wrote his famous aphorism. If Wiles is right in his opinion that Fermat did not have a proof of his conjecture then the Pythagorean approach may well be nugatory. However, a close examination of the relationship between

 A² + B² = C² and aⁿ + bⁿ = cⁿ when n > 2 provides valuable insights into the nature of the problem.      

Fermat’s conjecture can be stated as follows:

There are no positive integers a,b,c,n such that aⁿ + bⁿ = cⁿ given that  n > 2 and that a,b and c have no common factor. It follows from this last condition that a,b and c must be distinct integers. Formally:

FLT = _df NSabcnK² aⁿ + bⁿ = cⁿ a, b, c, n ∈ ℕ₁  a,b,c HNCF

It is convenient to specify that a < b, since this does not affect the argument, and it is easy to show that a and c must be odd integers and b must be an even integer under the given conditions.

The definition of the conjecture suggests a strategy for its proof. The contrary of the conjecture NFLT can only be true if two conditions are jointly satisfied: the equality must be true and all three variables that a,b and c must be integers. The aim, therefore, is to show that one of these conditions must be false.  Regarding the second condition, it is only necessary to show that one of the variables  a,b and c must be irrational under the given conditions. It should be noted that the rational fractional case is subsumed in the integral case.  

The key to the Pythagorean approach is the identity: (a^n/2)²  =_ID   aⁿ and the definition

A = a^n/2, the consequence being that A² =_ID aⁿ. The application of this principle to each term of the Fermat equation produces the identity aⁿ + bⁿ = cⁿ =_ID  A² + B² = C². The identity symbol signifies more than the logical equivalence of the two expressions since it also requires that any rearrangement of one side must be identical to that of the other. For the purposes of the proof the definitions used are as follows: D₁:  aⁿ = A², bⁿ = B², cⁿ = C².

The purpose of deploying the Pythagorean equation is that the conditions for its truth or falsity can be perfectly defined. The identity between the equations provides the means of knowing the truth or falsity of the equality in the Fermat equation. Only the implications of the truth of aⁿ + bⁿ = cⁿ are relevant since the falsity of the equation would lead immediately to the assertion of FLT.

The condition for the truth of A² + B² = C² is given by the following identity:

A² + B² = C²  =_ID  (x² - y²)² + 4x²y² = (x² + y²)² under the definitions D₂: A = x² - y², B = 2xy, C = x² + y².

Since aⁿ + bⁿ = cⁿ =_ID  A² + B² = C² it follows that aⁿ + bⁿ = cⁿ =_ID (x² - y²)² + 4x²y² = (x² + y²)².

The consequence is that aⁿ = (x² - y²)², bⁿ =4x²y² and cⁿ = (x² + y²)².  This provides a model for examining the implication of aⁿ + bⁿ = cⁿ a, b, c, n ∈ ℕ₁  a,b,c HNCF = _df NFLT.

The equality bⁿ =4x²y² is the most tractable for determining the status of the variable b.  Since bⁿ is an even number b must also be even. The equation (x² - y²)² + 4x²y² = (x² + y²)² is true for all real values of x and y and consequently Pxy (x² - y²)² + 4x²y² = (x² + y²)² x,y ∈ ℕ₁  x,y HNCF is a theorem.

The expression 4x²y² can only have an nth root when 4x²y² = 2ⁿuⁿvⁿ, where u,v HNCF, consequently

bⁿ = 2ⁿuⁿvⁿ, and so the definition of b becomes  D₃:  b = 2uv. The final definition is D₄: vⁿ = y², where y and v are odd numbers. These definitions and key propositions are listed below.

Before proceeding to the formal proof of it is useful to analyse A² + B² = C² in terms of the derived variables u and v. The rationality or otherwise of the variables can then be determined.

           

A2	B2	C2
an	bn	cn
(x2 - y2)2	4x2y2	(x2 + y2)2
(2n-2un - vn)2	2nunvn	(2n-2un + vn)2

If 2ⁿuⁿvⁿ = 4x²y² then uⁿ = 4x²y² / 2ⁿvⁿ but vⁿ = y² so uⁿ  = 4 x² / 2ⁿ and  u = x^2/n/ 2^(n-2)/n. It is now clear that u is irrational under the given condition. This is true whether or not x is rational. This is relevant because if n ∤ 2 then B is either an irrational square root or is a square. If B =  x² - y²  is irrational  then b^n/2 is an irrational square root unless b is a square.

The significant conclusion is that while aⁿ + bⁿ = cⁿ and aⁿ, bⁿ , cⁿ ∈ ℕ₁  the variable b is irrational.

But if b ∉ ℚ then b ∉ ℕ₁ . The irrationality of b is inconsistent with the condition that b must be a positive integer as required by the condition on both FLT and NFLT. This logical contradiction has the form CpNp , which combined with the theorem CCpNPNp allows the final conclusion that NFLT is false and so FLT must be true.

The elementary nature of the mathematics and logic involved suggests that Fermat might well have arrived at the same conclusion, but neglected to write it down. The use of a variant of the theorem

CCpCNpp (consequentia mirabilis) could explain his claim that he had discovered a marvellous proof. The marginalia he could have squeezed in might then have been as follows:

The supposition aⁿ + bⁿ = cⁿ , where a, b and c are integers, implies that b is irrational and so  contradicts itself. The contrary proposition, therefore, must be true, consequentia mirabilis.

Definitions

D₁:       aⁿ = A², bⁿ = B², cⁿ = C²

D₂:       A = x² - y², B = 2xy, C = x² + y²

D₃:       b = 2uv

D₄:       vⁿ = y²

Proposition                                                                Condition

T₁:        SABC A² + B² = C²                                 A,B,C ∈ ℕ₁ A,B,C HNCF

T₂:        Pxy (x² - y²)² + 4x²y² = (x² + y²)²           x,y ∈ ℕ₁  x,y HNCF

P₁:        Sabcn aⁿ + bⁿ = cⁿ                                CondP₁: n > 2 a, b, c, n ∈ ℕ₁  a,b,c HNCF

Formal Proof:  NSabcnK² aⁿ + bⁿ = cⁿ a, b, c, n ∈ ℕ₁ a,b,c HNCF

Line	Step		Justification
1	an = A2, bn = B2, cn = C2	D1	df in.
2	A = x2 - y2, B = 2xy, C = x2 + y2	D2	df in.
3	b = 2uv	D3	df in.
4	vn = y2	D4	df in.
5	SABC A2 + B2 = C2 (A,B,C Î ℕ1 ,  A,B,C HNCF)	T1	Th in.
6	Pxy (x2 - y2)2 + 4x2y2 = (x2 + y2)2 (x,y ∈ ℕ1 ,  x,y HNCF)	T2	Th in
7	Sabcn an + bn = cn o  (n > 2, a, b, c, n ∈ ℕ1 , a,b,c HNCF)	P1	Supposition.
8	bn = 4x2y2	P2	1,2 defs D1, D2
9	bn = 2nunvn	P3	1,3 defs D1, D3
10	2nunvn = 4x2y2	P4	8,9 equality
11	un  = 4x2y2 / 2nvn	P5	10 rearrange
12	un  = 4 x2 / 2n	P6	11 def 4
13	u = x2/n/ 2(n-2)/n	P7	12 rearrange
14	u ∉ ℚ	P8	13 u is irrational
15	C u ∉ ℚ b ∉ ℚ	P9	14 3, D3
16	b ∉ ℚ	P10	14, 15 MP C exp.
17	C b ∉ ℚ b ∉ ℕ1	P11	16 def rational
18	b ∉ ℕ1	P12	15.16 MP C exp.
19	C P1 b ∈ ℕ1	P13	7, P1 conditions
20	NP1	P14	18, 19 MD C exp.
21	C P1 NP1	P15	7,20 C in.
22	CC P1 NP1 NP1	P16	th in. (CCpqq)
23	C P1NP1   = P15	P17	21 def in.
24	CP15 NP1	P18	22,23 def P15
25	P15	P19	21 is P15
26	KCP15 NP1P15	P20	24, 25 K in.
27	NP1	P21	26 MP C exp.
28	NSabcn an + bn = cn  (n > 2, a, b, c, n ∈ ℕ1 , a,b,c HNCF)	P22	7 refuted RA.

Tony Thomas ©
October 2012