Friday, December 14, 2012

Fermat’s Conjecture Revisited



The Application of Pythagorean Triples to Fermat’s Conjecture
a Parametric Approach
Tony Thomas
Abstract
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 1, 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 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.

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  =ID  an and the definition A = an/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 1  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 1  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 =
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 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 q2 and where p2= 2k, producing the expression 4x2y2= 2k+2q2y2. Since the odd factors can have nth roots the deciding factor is how 2k+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 = 2m so that m = log2b – log2s – log2v. 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 of n.  If all the conditions are satisfied, an = (2k – 1)2 and cn = (2k + 1)2. The remainder of the proof shows that (2k – 1)2and (2k + 1)2 cannot both have rational nth roots. The consequence is that the condition a, b, c, n Î 1 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 1 A,B,C HNCF
T2:        Pxy (x2 - y2)2 + 4x2y2 = (x2 + y2)2        x,y Î 1  x,y HNCF, x|2, y 2
Q:        Sabcn an + bn = cn                               Qc:  n > 2 a, b, c, n Î 1  a,b,c HNCF

Lemma 1: C n 2 x
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
5 fractional index
7
x = 2k/2q
3 square root
8
C n 2 x
6,7

Proof 1:  C n 2 FLT
Line
Step
Justification
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 x Î 1
 c2 condition on x
8
 x Î 1
 6,7 MP C exp.
9
x Î
8 C1
10
      x Î
9 tr.
11
     K x Î   x
3,10 K in.
12
C K n 2  P1 K x Î   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]
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
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 2 o
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
C 2log2x ℚ k ∉
4
15
      C 2log2x ℚ k ∉
12, tr.
16
     k ∉
12, 15 C exp.
17
C n 2 k ∉
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 n

3 irrationality of both sides
5
EA2 b ≠ 2l s ≠ 1 v ≠ 1 m

1 rationality of logs
6
EA2 b ≠ 2l s ≠ 1 v ≠ 1 n

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

6, 7
20
C n Î ℚ S1

19 CEpqCpq
21
C n Î K c = (2k +1)2/n a = (2k  - 1)2/n

18, 20 trans.

Lemma 5:  NSfgK fn – gn = 2 K f Î 1  g Î 1
Line
Step

Justification
1
      SfgKfn –gn = 2 Kf Î 1 g Î 1o
P1

2
       fg|Kfn –gn = 2 Kf Î 1 g Î
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 1  g Î
P17
16 expand

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

Justification
1
K(2k +1)1/n Î (2k – 1)1/n Î o
P1
Supposition
2
(2k +1)1/n =  SfK 2k +1= fn fÎ1 
P2
d10 df in.
3
(2k - 1)1/n =  SfK 2k - 1= gn fÎ1 
P3
d10 df in.
4
      (2k +1)1/n
P4
1 K exp.
5
      (2k – 1)1/n
P5
1 K exp.
6
    SfK 2k +1= fn fÎ1 
P6
2, 4 df sub.
7
    SgK 2k - 1= gn fÎ1 
P7
3.5 df sub
8
           f | K 2k +1= fn fÎ1 
P8
6 S exp.
9
           g| K 2k - 1= gn gÎ1 
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
           fÎ1 
P13
8 K exp.
14
          gÎ1 
P14
9 K exp.
15
          K fÎ1  gÎ1 
P15
13, 14 K in.
16
          K fn – gn = 2K fÎ1  gÎ1 
P16
12, 15 K in.
17
      SfgK fn – gn = 2K fÎ1  gÎ1 
P17
16 S in.
18
CP1 SfgK fn – gn = 2K fÎ1  gÎ1 
P18
1, 17 C in.
19
NSfgK fn – gn = 2K fÎ1  gÎ1 
P19
Lemma 5
20
NP1
P20
18, 19 C exp.
21
NK(2k +1)1/n Î (2k – 1)1/n Î
P21
20 NP1 expand
22
A(2k +1)1/n   (2k – 1)1/n  
P22
21 De Morgan




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

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

Proof 2: NKSabcn an +bn = cn K2 n > 2 a, b, c, n Î 1 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 1 a, b, c HNCF
P2
Qc  df in
3
      Q o
P3
Supposition
4
      Qc
P4
1 K exp.
5
      K3 a Î1  b Î1  c Î1  n Î1 
P5
2, 4 expand K exp.
6
      C n Î A c   ℚ a ∉  
P6
Lemma 8
7
     n Î 1
P7
5 K exp.
8
      n
P8
7 C 1
9
      A c ℚ a  
P9
6, 8 C exp.
10
      K c Î 1  a Î 1 
P10
5 K exp.
11
      A c 1  a   1 
P11
9 C x ℚ x 1  
12
      N K c Î 1  a Î 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.
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= 2k
d7:  b = 2uv
d8: vn = y2
d9: fn = 2k + 1, gn = 2k - 1
Conditions
c1: A, B, C Î 1 ,  A, B, C  HNCF
c2: a, b, c,  Î 1 ,  a, b, c  HNCF, a 2, c 2, b | 2
c3: n Î 1, n > 2 
c4: x, y Î 1 , x, y  HNCF , x | 2, y  2
c5: u, v Î 1 , u, v  HNCF , u | 2, v 2
c6: p|2 only,  q 2    
c7: r|2 only, s 2
c8: l, m Î 1
c9: f, g Î