fermat's conjecture revisited

7/29/2019 Fermat's Conjecture Revisited

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The Application of Pythagorean Triples to Fermats Conjecture

a Parametric Approach

Tony Thomas

Abstract

This paper applies the theory of Pythagorean triples to Fermats Conjecture, the contrary of

which requires that an

+ bn

= cn

where a, b, c, n 1, n>2. Firstly, it is proven, for odd

values ofn, that the parameterx of the Pythagorean triples must be irrational under the

negation of Fermats 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 aproof of Fermats 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

Fermats age but the formal logic used does exceed the Aristotelian syllogistic of his time.

The identity betweenA2

+ B2

= C2 and a

n+ b

n= c

n when n > 2 provides valuable insights

into the nature of the problem. Fermats 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 = dfNSabcnK2

an

+ bn

= cn

a, b, c, n 1 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.


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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: (an/2

)2

=IDa

nand the definition A = a

n/2,

the consequence being thatA2

=ID an. The application of this principle to each term of the

Fermat equation produces the identity an

+ bn

= cn

=IDA2

+ 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, b

n= B

2, c

n= C

2.

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 ofan

+ bn

= cnare relevant since the falsity of the equation would

lead immediately to the assertion of FLT.

The condition for the truth ofA2

+ B2

= C2

is given by the following identity:

A2

+ B2

= C2 =ID (x

2- y

2)2

+ 4x2y

2= (x

2+ y

2)2 under the definitions d2: A =x

2- y

2, B = 2xy,

C = x2

+ y2. Since a

n+ b

n= c

n=IDA

2+ B

2= C

2it follows that a

n+ b

n= c

n=ID (x

2- y

2)2

+

4x2y

2= (x

2+ y

2)2. The consequence is that a

n= (x

2- y

2)2, b

n=4x

2y

2and c

n= (x

2+ y

2)2. This

provides a model for examining the implication ofan

+ bn

= cn

a, b, c, n 1 a,b,c HNCF =

dfNFLT.

The equality bn

=4x2y

2is the most tractable for determining the status of the variable b. The

equation (x2

- y2)2

+ 4x2y

2= (x

2+ y

2)2

is true for all real values ofx andy and consequently

Sxy (x2

- y2)2

+ 4x2y

2= (x

2+ y

2)2x,y 1 x,y HNCFis also a theorem. The expression 4x

2y

2

can only have an nth root when 4x2y

2= 2

nu

nv

n, where u and v have no common factors,

consequently bn

= 2nu

nv

n, and so the definition ofb becomes d3: b = 2uv. The remaining

definition is d4: vn = y2, wherey and v are odd numbers. The case for this rests on the


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assumption that v andy are indivisible by 2. The formx2

+ 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 analyseA2

+ 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 B

2 C

2

an b

n c

n

(x2

- y2)2 4x

2y

2 (x

2+ y

2)2

( 2n-2

un

- vn

)2

2n

un

vn

(2n-2

un

+ vn

)2

(2kq

2v

n)2 2

k+2q

2v

n(2

kq

2+ v

n)2

(2k1)

22

k+2(2k + 1)

2

If2nu

nv

n= 4x

2y

2thenu

n= 4x

2y

2/ 2

nv

nbut v

n= y

2so u

n = 4 x

2/ 2

nand u =

x2/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 thatx 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 ifq = 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

= 4x2y

2andresolving the question as to how the RHS

can have a rational nth rootwhen n > 2. The proof divides 4x2y

2into odd and even factors:

4x2y

2=4p

2q

2, where p|2 only and q2 and wherep

2= 2

k, producing the expression 4x

2y

2=

2k+2

q2y

2. Since the odd factors can have nth roots the deciding factor is how 2

k+2could have

an nth root. This is only possible when mn = k + 2, where m and kare dummy variables. It

follows from this definition that m and kmust have the same parity as n, and that the


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rationality ofn depends on the rationality ofkand m. The proof shows that b/sv = 2mso that

m = log2blog2slog2v. The consequence is that m can only be rational when s = v = 1 and

b = 2m. If none of these conditions is satisfied then n is irrational and FLT immediately

follows from the irrationality ofn. If all the conditions are satisfied, an

= (2k1)

2and c

n=

(2k

+ 1)2. The remainder of the proof shows that (2

k1)

2and (2

k+ 1)

2cannot both have

rational nth roots. The consequence is that the condition a, b, c, n 1cannot be true and

FLT must be true. In proof 2, FLT is redefined asNKSabcn 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 ofJakowski 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,C1 A,B,C HNCF

T2: Pxy (x2

- y2)2

+ 4x2y

2= (x

2+ y

2)2

x,y 1 x,y HNCF, x|2, y 2

Q: Sabcn an + bn = cn Qc: n > 2 a, b, c, n 1 a,b,c HNCF

Lemma 1: C n 2 x

Line Step Justification

1 x2

= p2q

2df in. d1

2 p2= 2

kdf in. d6

3 x2

= 2kq

21,2

4 k = mn2 df in. d5

5 C n 2 k2 4

6 C n 2 2k/2

5 fractional index

7 x = 2k/2

q 3 square root

8 C n 2 x 6,7


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Proof 1: C n 2 FLT


1 CK n 2 P1 x Lemma 1

2 K n 2 P1o Supposition

3 x 1.2 MP C exp.

4 T2 Theorem in.

5 CT2c2 T2 condition set

6 c2 4,5 MP Cexp.

7 Cc2 x1 c2 condition on x

8 x1 6,7 MP C exp.

9 x 8 C1

10 x 9 tr.

11 K x x 3,10 K in.

12 C K n 2 P1 K x x 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 [ancillary lemma not used in proof 2]


1 x2

= 2kq

2Lemma 1

2 2k

= x2/q

2 rearrange

3 log22k= log2(x2/q2) Take logs base 2


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4 k = 2log2x2log2q Evaluate logs

5 CK q > 1 q 2 log2q Log of odd number

6 q 2 C6 cond. On q

7 C q > 1 log2q 5, 6

8 C n 2 x Lemma 1

9 n 2o Supposition

10 x 8 tr. 9 C exp.

11 C x 2log2x Log of irrational #

12 2log2x 10, 11 C exp.

13 C n 2 2log2x 9, 12 C in.

14 C2log2x k 4

15 C2log2x k 12, tr.

16 k 12, 15 C exp.

17 C n 2 k 9, 16 C in.

Lemma 3:m = log2blog2slog2v


1 b = 2uv d7 df in.

2 u = rs d4 df in.

3 un

= 4x2/2

n d1, d2 dfs in.

4 x2

= 2kq

2 Lemma 1

5 un

=2k+2

q2/2

n3, 4

6 rns

n=2

k+2q

2/2

n2, 5

7 2nr

n/2

k+2= q

2/s

n6 parity error

8 2nr

n= 2

k+2 7 PE resolved


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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

= log2blog2slog2v 17 take logs base 2

19 m = log2blog2slog2v evaluate

Lemma 4: C n


1 m = log2blog2slog2v Lemma 3

2 mn2 = k D5 df in.

3 n = (k + 2)/m 2 rearrange

4 E m n 3 irrationality of

5 EA2

b 2

s 1 v 1 m 1 rationality of

6 EA2

b 2

s 1 v 1 n 4, 5 E subs.

7 K2

b = 2

s = 1 v = 1o S1 Supposition

8 x2

= 2kq

2 Lemma 1

9 q2

= sn Lemma 3

10 q2

= 1 7 supposition


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11 x2

+ y2

= 2k+1 7, 8, 10

12 x2

- y2

= 2k- 1 7, 8, 10

13 (x

2

+ y

2

)

2

= c

n

d1, d2 defs in.

14 (x2y

2)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/na = (2

k- 1)

2/n 15, 16 K in.

18 CS1 K c = (2k+1)

2/na = (2

k- 1)

2/n 7, 17 C in.

19 ES1 n 6, 7

20 C n S1 19 CEpqCpq

21 C n K c = (2k+1)

2/na = (2

k- 1)

2/n 18, 20 trans.

Lemma 5: NSfgK fn

gn

= 2 K f 1g 1


1 SfgKfng

n= 2 Kf1 g 1

o P1

2 fg|Kfng

n= 2 Kf1 g P2 1 S exp.

3 fng

n= 2 P3 2 K exp.

4 fn

= 2k

+ 1 P4 d9 df in.

5 gn

= 2k

- 1 P5 d9 df in.

6 f2 P6 4 RHS odd

7 g2 P7 5 RHS odd

8 n > 2 P8 c2 in.

9 K2

f2 g2 n > 2 P9 6, 7, 8 K2

in.

10 C P9 fng

n> 2 P10 9 f >g. g > 1


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11 fng

n> 2 P11 9, 10 C exp.

12 fng

n 2 P12 11

13 K f

n

g

n

= 2 f

n

g

n

2 P13 3, 12 K in.

14 Sfg K fng

n= 2 f

ng

n 2 P14 1. S in.

15 C P1K P12N P12 P15 14

16 N P1 P16 15 C exp.

17 NSfgK fng

n= 2 K f1 g P17 16 expand

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


1 K(2k

+1)1/n (2k1)

1/n

o P1 Supposition

2 (2k

+1)1/n

= SfK 2k

+1= fn

f1 P2 d10 df in.

3 (2k

- 1)1/n

= SfK 2k

- 1= gn

f1 P3 d10 df in.

4 (2k

+1)1/n

P4 1 K exp.

5 (2k1)1/n

P5 1 K exp.

6 SfK 2k

+1= fn

f1 P6 2, 4 df sub.

7 SgK 2k

- 1= gn

f1 P7 3.5 df sub

8 f|K 2k

+1= fn

f1 P8 6 S exp.

9 g| K 2k

- 1= gn

g1 P9 7 S exp.

10 2k

+1= fn P10 8 K exp.

11 2k

- 1= gn P11 9 K exp

12 fng

n= 2 P12 10, 11 diff

13 f1 P13 8 K exp.

14 g1 P14 9 K exp.

15 K f1 g1 P15 13, 14 K in.


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16 K fng

n= 2K f1 g1 P16 12, 15 K in.

17 SfgK fng

n= 2K f1 g1 P17 16 S in.

18 CP1 SfgK f

n

g

n

= 2K f1 g1 P18 1, 17 C in.

19 NSfgK fng

n= 2K f1 g1 P19 Lemma 5

20 NP1 P20 18, 19 C exp.

21 NK(2k

+1)1/n (2k1)

1/n P21 20 NP1 expand

22 A(2k

+1)1/n

(2k1)1/n

P22 21 De Morgan

Lemma 7: E w1/n w2/n


1 n > 2o Supposition

2 C w1/n

w2/n 1 2/n < 1

3 Cw2/n w1/n

1 1/n < 1

4 E w1/n

w2/n 2, 3 E in.

5 C n > 2 E w1/n

w2/n 1, 4 C in.

6 n > 2 C2 condition

7 E w1/n

w2/n 5, 6 C exp.

Lemma 8: C n


1 E (2k

+1)1/n

(2k1)2/n

Lemma 7 subs.

2 A (2k

+1)2/n

(2k1)2/n

Lemmas 6, 7

3 no Suppostion

4 K c = (2k

+1)2/n

a = (2k1)2/n

Lemma 4

5 c = (2k+1) 2/n 4 K exp.


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6 a = (2k1)2/n 4 K exp.

7 A (2k

+1)2/n

(2k1)2/n

2 tr.

8 A c a 5, 6, 7 subs.

9 C nA c a 3, 8 C exp.

Proof 2: NKSabcn an

+bn

= cn

K2n > 2 a, b, c, n 1a, b, c HNCF


1 Q = df KSabcn an

+bn

= cn

Qc P1 Q df in.

2Qc = df K

2

n > 2 a, b, c, n 1 a, b, c

HNCFP2 Qc df in

3 Qo P3 Supposition

4 Qc P4 1 K exp.

5 K3

a 1 b 1 c 1 n 1 P5 2, 4 expand K exp.

6 C nA c a P6 Lemma 8

7 n 1 P7 5 K exp.

8 n P8 7 C1

9 A c a P9 6, 8 C exp.

10 K c 1 a 1 P10 5 K exp.

11 A c 1 a 1 P11 9 Cx x1

12 N K c 1 a 1 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.


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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, whereA2

+ 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 ofx andy

p, q, r, s : parameters ofu and v

parameters oflog x and log y

Definitions

d1: a

n

= A

2

, b

n

= B

2

, c

n

= C

2

d2:A =x2

- y2, B = 2xy, C = x

2+ y

2

d3:x2

= p2q

2

d4: u = rs

d5: k = mn2

d6:p2= 2k

18 FLT P18 17 df sub.


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d7: b = 2uv

d8: vn

= y2

d9:f

n

= 2

k

+ 1, g

n

= 2

k

- 1

Conditions

c1:A, B, C1, A, B, C HNCF

c2: a, b, c, 1, a, b, c HNCF, a 2, c 2, b | 2

c3: n 1, n > 2

c4:x, y1, x, y HNCF , x | 2, y 2

c5: u, v1, u, v HNCF , u | 2, v 2

c6:p|2 only, q 2

c7: r|2 only, s 2

c1

c9:f, g

fermat's conjecture revisited

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