Fermat’s Conjecture Revisited

The Application of Pythagorean Triples to Fermat’s Conjecture

a Parametric Approach

Tony Thomas

Abstract

[.pdf here]

This paper applies the theory of Pythagorean triples to Fermat’s Conjecture, the contrary of which requires that aⁿ + bⁿ = cⁿ where a, b, c, n ∈ ℕ₁, n>2.  Firstly, it is proven, for odd values of n, that the parameter x of the Pythagorean triples must be irrational under the negation of Fermat’s Conjecture (NFLT), the consequence being that the conjecture is true. Secondly, it is proven that either the variables a and c are irrational or n itself is irrational under the negative conjecture. The truth of FLT follows from this.

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. The proofs in this paper involve no mathematical ideas not current in Fermat’s age but the formal logic used does exceed the Aristotelian syllogistic of his time.

The identity 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 both a and c must be odd integers and that 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 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 implies that the corresponding terms are identical. 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 by a theorem. 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.  The equation (x² - y²)² + 4x²y² = (x² + y²)² is true for all real values of x and y and consequently Sxy (x² - y²)² + 4x²y² = (x² + y²)²  x,y Î ℕ₁  x,y HNCF is also a theorem. The expression 4x²y² can only have an nth root when 4x²y² = 2ⁿuⁿvⁿ, where u and v have no common factors, consequently bⁿ = 2ⁿuⁿvⁿ, and so the definition of b becomes d₃:  b = 2uv. The remaining definition is d₄: vⁿ = y², where y and v are odd numbers. The case for this rests on the assumption that v and y are indivisible by 2. The form x² + y² must comprise even and odd squares respectively and so the second term must be odd. This is only possible when the second term is vⁿ rather than 2vⁿ . These definitions and key propositions are listed below.

Before proceeding to the formal proofs 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.

Table of Identities

A2	B2	C2
an	bn	cn
(x2 - y2)2	4x2y2	(x2 + y2)2
( 2n-2un - vn)2	2nunvn	(2n-2un + vn)2
(2kq2 – vn )2	2 k+2q2vn	(2kq2 + vn )2
(2k – 1)2	2 k+2	(2k + 1)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 may appear that u is irrational under the given conditions, however, this can be challenged by positing dummy variables m and n, defined in d5 of the Annex. The consequence is that all direct attempts to prove b is irrational fail.

Proofs

Lemma 1 provides the basis for the first proof, and shows that x cannot be rational under the assumptions of NFLT.  This contradicts Theorem 2 and so refutes NFLT. The second proof examines the dummy variables and finds them to be irrational; the consequence being the refutation of NFLT on the grounds that n must also be rational under the given constraints. However, n can still be an integer if q = v =1, but this results in the irrationality of either a or c and so leads to the refutation of NFLT.

Outline of Proof 2

The key to this proof is the relation bⁿ = 4x²y²and resolving the question as to how the RHS can have a rational nth rootwhen n > 2. The proof divides 4x²y²  into odd and even factors:  4x²y²=4p²q², where p|2 only and q∤ 2 and where p²= 2^k, producing the expression 4x²y²= 2^k+2q²y². Since the odd factors can have nth roots the deciding factor is how 2^k+2 could have an nth root. This is only possible when mn = k + 2, where m and k are dummy variables. It follows from this definition that m and k must have the same parity as n, and that the rationality of n depends on the rationality of k and m. The proof shows that b/sv = 2^m so that m = log₂b – log₂s – log₂v. The consequence is that m can only be rational when s = v = 1 and b = 2^m. If none of these conditions is satisfied then n is irrational and FLT immediately follows from the irrationality of n.  If all the conditions are satisfied, aⁿ = (2^k – 1)² and cⁿ = (2^k + 1)². The remainder of the proof shows that (2^k – 1)²and (2^k + 1)² cannot both have rational nth roots. The consequence is that the condition a, b, c, n Î ℕ₁ cannot be true and FLT must be true. In proof 2, FLT is redefined as NKSabcn aⁿ +bⁿ = cⁿ Qc. The corollary is that CQcNSabcn aⁿ +bⁿ = cⁿ .  In other words, if the given conditions apply then FLT is true.

Definitions of the variables and the conditions applying to them are set out in the Annex and are referred to in the proofs by their item numbers. The proofs follow the suppositional method of Jaśkowski and Gentzen and are expressed in Polish notation. Details of the method are expounded in the text by P H Nidditch [1].

Proposition                                                     Condition

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

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

Q:        Sabcn aⁿ + bⁿ = cⁿ                               Qc:  n > 2 a, b, c, n Î ℕ₁  a,b,c HNCF

Lemma 1: C n ∤ 2 x ∉ ℚ

Line	Step	Justification
1	x2 = p2q2	df in. d1
2	p2= 2k	df in. d6
3	x2 = 2kq2	1,2
4	k = mn – 2	df in. d5
5	C n ∤ 2 k ∤ 2	4
6	C n ∤ 2  2 k/2 ∉ ℚ	5 fractional index
7	x = 2k/2q	3 square root
8	C n ∤ 2 x ∉ ℚ	6,7

Proof 1:  C n ∤ 2 FLT

Line	Step	Justification
1	CK n ∤ 2  P1 x ∉ ℚ	Lemma 1
2	K n ∤ 2  P1o	Supposition
3	x ∉ ℚ	1.2 MP C exp.
4	T2	Theorem in.
5	CT2c2	T2 condition set
6	c2	4,5 MP Cexp.
7	Cc2 x Î ℕ1	c2 condition on x
8	x Î ℕ1	6,7 MP C exp.
9	x Î ℚ	8 Cℕ1ℚ
10	x Î ℚ	9 tr.
11	K x Î ℚ  x ∉ ℚ	3,10 K in.
12	C K n ∤ 2  P1 K x Î ℚ  x ∉ ℚ	3.11 C in.
13	N K n ∤ 2 P1	12 ENpCpKqNq
14	C n ∤ 2  NP1	13 De Morgan
15	P1 = df FLT	df in.
16	C n ∤ 2 FLT	14,15

Lemma 2: C n ∤ 2 k ∉ ℚ  [ancillary lemma not used in proof 2]

Line	Step	Justification
1	x2 = 2kq2	Lemma 1
2	2k = x2/q2	rearrange
3	log22k = log2(x2/q2)	Take logs base 2
4	k = 2log2x – 2log2q	Evaluate logs
5	CK q > 1 q ∤ 2 log2q ∉ ℚ	Log of odd number
6	q ∤ 2	C6 cond. On q
7	C q > 1 log2q ∉ ℚ	5, 6
8	C n ∤ 2 x ∉ ℚ	Lemma 1
9	n ∤ 2 o	Supposition
10	x ∉ ℚ	8 tr. 9 C exp.
11	C x ∉ ℚ 2log2x ∉ ℚ	Log of irrational #
12	2log2x ∉ ℚ	10, 11 C exp.
13	C n ∤ 2 2log2x ∉ ℚ	9, 12 C in.
14	C 2log2x ∉ ℚ k ∉ ℚ	4
15	C 2log2x ∉ ℚ k ∉ ℚ	12, tr.
16	k ∉ ℚ	12, 15 C exp.
17	C n ∤ 2 k ∉ ℚ	9, 16 C in.

Lemma 3: m = log₂b – log₂s – log₂v

Line	Step	Justification
1	b = 2uv	d7 df in.
2	u = rs	d4 df in.
3	un = 4x2/2n	d1, d2 dfs in.
4	x2 = 2kq2	Lemma 1
5	un =2k+2 q2/2n	3, 4
6	rnsn = 2k+2 q2/2n	2, 5
7	2nrn/2 k+2  = q2/sn	6 parity error
8	2nrn = 2k+2	7 PE resolved
9	q2 = sn	7 PE resolved
10	rn = 2k+2- n	8 rearrange
11	k = mn - 2	D5 df in.
12	rn =2mn - n	10, 11
13	r = 2m-1	12 nth root
14	2r = 2m	13
15	b = 2rsv	1, 2
16	b = 2msv	14, 15
17	2m = b/sv	16 rearrange
18	log22m = log2b – log2s – log2v	17 take logs base 2
19	m = log2b – log2s – log2v	evaluate

Lemma 4: C n Î ℚ K c = (2k + 1)^2/n a = (2k – 1)^2/n

Line	Step		Justification
1	m = log2b – log2s – log2v		Lemma 3
2	mn – 2 = k		D5 df in.
3	n = (k + 2)/m		2 rearrange
4	E m ∉ ℚ n ∉ ℚ		3 irrationality of both sides
5	EA2 b ≠ 2l s ≠ 1 v ≠ 1 m ∉ ℚ		1 rationality of logs
6	EA2 b ≠ 2l s ≠ 1 v ≠ 1 n ∉ ℚ		4, 5 E subs.
7	K2 b = 2l  s = 1 v = 1 o	S1	Supposition
8	x2 = 2kq2		Lemma 1
9	q2 = sn		Lemma 3
10	q2 = 1		7 supposition
11	x2 + y2 = 2k +1		7, 8, 10
12	x2 - y2 = 2k - 1		7, 8, 10
13	(x2 + y2 )2= cn		d1, d2 defs in.
14	(x2 – y2)2 = an		d1, d2 defs in
15	c = (2k +1)2/n		13 nth root
16	a = (2k  - 1)2/n		14 nth root
17	K c = (2k +1)2/n a = (2k  - 1)2/n		15, 16 K in.
18	CS1  K c = (2k +1)2/n a = (2k  - 1)2/n		7,  17 C in.
19	ES1  n Î ℚ		6, 7
20	C n Î ℚ S1		19 CEpqCpq
21	C n Î ℚ K c = (2k +1)2/n a = (2k  - 1)2/n		18, 20 trans.

Lemma 5:  NSfgK fⁿ – gⁿ = 2 K f Î ℕ₁  g Î ℕ₁

Line	Step		Justification
1	SfgKfn –gn = 2 Kf Î ℕ1 g Î ℕ1o	P1
2	fg|Kfn –gn = 2 Kf Î ℕ1 g Î ℕ	P2	1 S exp.
3	fn –gn = 2	P3	2 K exp.
4	fn = 2k + 1	P4	d9 df in.
5	gn = 2k - 1	P5	d9 df in.
6	f ∤ 2	P6	4 RHS odd
7	g ∤ 2	P7	5 RHS odd
8	n > 2	P8	c2 in.
9	K2 f ∤ 2 g ∤ 2 n > 2	P9	6, 7, 8 K2 in.
10	C P9 fn –gn > 2	P10	9 f >g. g > 1
11	fn –gn > 2	P11	9, 10 C exp.
12	fn –gn  ≠ 2	P12	11
13	K fn –gn = 2 fn –gn  ≠ 2	P13	3, 12 K in.
14	Sfg K fn –gn = 2 fn –gn  ≠ 2	P14	1. S in.
15	C P1K P12N P12	P15	14
16	N P1	P16	15 C exp.
17	NSfgK fn –gn = 2 K f Î ℕ1  g Î ℕ	P17	16 expand

Lemma 6:  A(2^k + 1)^1/n ∉ ℚ (2^k -1)^1/n ∉ ℚ

Line	Step		Justification
1	K(2k +1)1/n Î ℚ (2k – 1)1/n Î ℚ o	P1	Supposition
2	(2k +1)1/n ∉ ℚ =  SfK 2k +1= fn fÎℕ1	P2	d10 df in.
3	(2k - 1)1/n ∉ ℚ =  SfK 2k - 1= gn fÎℕ1	P3	d10 df in.
4	(2k +1)1/n ∉ ℚ	P4	1 K exp.
5	(2k – 1)1/n ∉ ℚ	P5	1 K exp.
6	SfK 2k +1= fn fÎℕ1	P6	2, 4 df sub.
7	SgK 2k - 1= gn fÎℕ1	P7	3.5 df sub
8	f | K 2k +1= fn fÎℕ1	P8	6 S exp.
9	g| K 2k - 1= gn gÎℕ1	P9	7 S exp.
10	2k +1= fn	P10	8 K exp.
11	2k - 1= gn	P11	9 K exp
12	fn – gn = 2	P12	10, 11 diff
13	fÎℕ1	P13	8 K exp.
14	gÎℕ1	P14	9 K exp.
15	K fÎℕ1  gÎℕ1	P15	13, 14 K in.
16	K fn – gn = 2K fÎℕ1  gÎℕ1	P16	12, 15 K in.
17	SfgK fn – gn = 2K fÎℕ1  gÎℕ1	P17	16 S in.
18	CP1 SfgK fn – gn = 2K fÎℕ1  gÎℕ1	P18	1, 17 C in.
19	NSfgK fn – gn = 2K fÎℕ1  gÎℕ1	P19	Lemma 5
20	NP1	P20	18, 19 C exp.
21	NK(2k +1)1/n Î ℚ (2k – 1)1/n Î ℚ	P21	20 NP1 expand
22	A(2k +1)1/n ∉  ℚ (2k – 1)1/n ∉  ℚ	P22	21 De Morgan

Lemma 7:  E w^1/n ∉ ℚ w^2/n  ∉  ℚ

Line	Step	Justification
1	n > 2 o	Supposition
2	C w1/n ∉  ℚ w2/n ∉  ℚ	1  2/n < 1
3	C w2/n ∉  ℚ w1/n ∉  ℚ	1  1/n < 1
4	E w1/n ∉  ℚ w2/n ∉  ℚ	2, 3 E in.
5	C n > 2 E w1/n ∉  ℚ w2/n ∉  ℚ	1, 4 C in.
6	n > 2	C2  condition
7	E w1/n ∉  ℚ w2/n ∉  ℚ	5, 6 C exp.

Lemma 8: C nÎ ℚ A c ∉  ℚ a ∉  ℚ

Line	Step	Justification
1	E (2k +1)1/n ∉  ℚ (2k – 1) 2/n ∉  ℚ	Lemma 7 subs.
2	A (2k +1) 2/n ∉  ℚ (2k – 1) 2/n ∉  ℚ	Lemmas 6, 7
3	n Î ℚ o	Suppostion
4	K c = (2k +1) 2/n  a = (2k – 1) 2/n	Lemma 4
5	c = (2k +1) 2/n	4 K exp.
6	a = (2k – 1) 2/n	4 K exp.
7	A (2k +1) 2/n ∉  ℚ (2k – 1) 2/n ∉  ℚ	2 tr.
8	A c ∉  ℚ a ∉  ℚ	5, 6, 7 subs.
9	C n Î ℚ A c ∉  ℚ a ∉  ℚ	3, 8 C exp.
Line	Step	Justification
1	E (2k +1)1/n ∉  ℚ (2k – 1) 2/n ∉  ℚ	Lemma 7 subs.
2	A (2k +1) 2/n ∉  ℚ (2k – 1) 2/n ∉  ℚ	Lemmas 6, 7
3	n Î ℚ o	Suppostion
4	K c = (2k +1) 2/n  a = (2k – 1) 2/n	Lemma 4
5	c = (2k +1) 2/n	4 K exp.
6	a = (2k – 1) 2/n	4 K exp.
7	A (2k +1) 2/n ∉  ℚ (2k – 1) 2/n ∉  ℚ	2 tr.
8	A c ∉  ℚ a ∉  ℚ	5, 6, 7 subs.
9	C n Î ℚ A c ∉  ℚ a ∉  ℚ	3, 8 C exp.

Proof 2: NKSabcn aⁿ +bⁿ = cⁿ K² n > 2 a, b, c, n Î ℕ₁ a, b, c HNCF

Line	Step		Justification
1	Q = df KSabcn an +bn = cn Qc	P1	Q  df in.
2	Qc = df K2 n > 2 a, b, c, n Î ℕ1 a, b, c HNCF	P2	Qc  df in
3	Q o	P3	Supposition
4	Qc	P4	1 K exp.
5	K3 a Îℕ1  b Îℕ1  c Îℕ1  n Îℕ1	P5	2, 4 expand K exp.
6	C n Î ℚ A c ∉  ℚ a ∉  ℚ	P6	Lemma 8
7	n Î ℕ1	P7	5 K exp.
8	n ∉ ℚ	P8	7 C ℕ1ℚ
9	A c ∉ ℚ a  ∉ ℚ	P9	6, 8 C exp.
10	K c Î ℕ1  a Î ℕ1	P10	5 K exp.
11	A c ∉ ℕ1  a  ∉ ℕ1	P11	9 C x ∉ ℚ x ∉ ℕ1
12	N K c Î ℕ1  a Î ℕ1	P12	10 De Morgan
13	K P10N P10	P13	10, 12   K in
14	CQ K P11N P11	P14	3, 13 C in.
15	NQ	P15	14  ECpNpNp
16	NKSabcn an +bn = cn Qc	P16	15 expand
17	FLT = df  NKSabcn an +bn = cn Qc	P17	16 df in.
18	FLT	P18	17 df sub.

References

[1] P. H. Nidditch, Introductory Formal Logic of Mathematics (University Tutorial Press, London 1957)

Tony Thomas

December 2012

Annex

Variables, definitions and conditions

Variables

A, B, C, where A² + B² = C²  : Pythagorean form

a, b, c, n, where aⁿ + bⁿ = cⁿ  : Fermat form

k, m : dummy variables

x, y : parameters of Pythagorean triples

u, v : parameters of x and y

p, q, r, s : parameters of u and v

l, m  : parameters of log x and log y

Definitions

d1: aⁿ = A², bⁿ = B², cⁿ = C²

d2: A = x² - y², B = 2xy, C = x² + y²

d3: x² = p²q²

d4: u = rs

d5: k = mn – 2

d6: p²= 2^k

d7:  b = 2uv

d8: vⁿ = y²

d9: fⁿ = 2^k + 1, gⁿ = 2^k - 1

Conditions

c1: A, B, C Î ℕ₁ ,  A, B, C  HNCF

c2: a, b, c,  Î ℕ₁ ,  a, b, c  HNCF, a ∤ 2, c ∤ 2, b | 2

c3: n Î ℕ₁, n > 2 

c4: x, y Î ℕ₁ , x, y  HNCF , x | 2, y ∤  2

c5: u, v Î ℕ₁ , u, v  HNCF , u | 2, v ∤ 2

c6: p|2 only,  q ∤ 2    

c7: r|2 only, s ∤ 2

c8: l, m Î ℕ₁

c9: f, g Î ℝ