The Application of Pythagorean
Triples to Fermat’s Conjecture
a Parametric Approach
Tony Thomas
This paper
applies the theory of Pythagorean triples to Fermat’s Conjecture, the contrary
of which requires that a^{n} + b^{n}
= c^{n} where a, b, c, n
∈ 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 A^{2
}+ B^{2} = C^{2} and a^{n} + b^{n} = c^{n} 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 a^{n }+
b^{n} = c^{n }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} NSabcnK^{2} a^{n}
+ b^{n} = c^{n} a, b, c, n ∈ 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: (a^{n/2})^{2 }=_{ID}^{ } a^{n} and the definition A = a^{n/2}, the consequence being
that A^{2 }=_{ID} a^{n}.
The application of this principle to each term of the Fermat equation
produces the identity a^{n} + b^{n}
= c^{n} =_{ID} A^{2 }+ B^{2} = C^{2}.
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: d_{1}: a^{n} = A^{2}, 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 of a^{n} + b^{n} =
c^{n} are relevant since the falsity of the equation would lead
immediately to the assertion of FLT.
The condition for the truth of A^{2 }+ B^{2} = C^{2} is given by the
following identity:
A^{2 }+
B^{2} = C^{2} =_{ID} (x^{2}
 y^{2})^{2} + 4x^{2}y^{2} = (x^{2} + y^{2})^{2}
under the definitions d_{2}: A =
x^{2}  y^{2}, B = 2xy, C
= x^{2} + y^{2}. Since a^{n}
+ b^{n} = c^{n} =_{ID} A^{2
}+ B^{2} = C^{2} it follows that a^{n} + b^{n} = c^{n} =_{ID }(x^{2}
 y^{2})^{2} + 4x^{2}y^{2} = (x^{2} + y^{2})^{2}.
The consequence is that a^{n}
= (x^{2}  y^{2})^{2}, b^{n} =4x^{2}y^{2}
and c^{n} = (x^{2} +
y^{2})^{2}. This
provides a model for examining the implication of a^{n} + b^{n} = c^{n} a, b, c, n ∈ ℕ_{1} a,b,c HNCF = _{df
}NFLT.
The equality b^{n}
=4x^{2}y^{2} is the most tractable for determining the
status of the variable b. The equation (x^{2}  y^{2})^{2} + 4x^{2}y^{2}
= (x^{2} + y^{2})^{2} is true for all real values
of x and y and consequently Sxy (x^{2}
 y^{2})^{2} + 4x^{2}y^{2} = (x^{2} + y^{2})^{2 }x,y Î ℕ_{1} x,y HNCF
is also a theorem. The expression 4x^{2}y^{2}
can only have an nth root when 4x^{2}y^{2
}= 2^{n}u^{n}v^{n}, where u and v have no common
factors, consequently b^{n} = 2^{n}u^{n}v^{n},
and so the definition of b becomes d_{3}: b = 2uv.
The remaining definition is d_{4}:
v^{n} = y^{2}, 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 x^{2} + y^{2} must
comprise even and odd squares respectively and so the second term must be odd.
This is only possible when the second term is v^{n} rather than 2v^{n}
. These definitions and key propositions are listed below.
Before proceeding to the formal proofs it is useful to
analyse A^{2 }+ B^{2} = C^{2}
in terms of the derived variables u
and v. The rationality or otherwise
of the variables can then be determined.
Table of Identities
A^{2}

B^{2}

C^{2}

a^{n}

b^{n}

c^{n}

(x^{2}  y^{2})^{2}

4x^{2}y^{2}

(x^{2} + y^{2})^{2}

( 2^{n2}u^{n}  v^{n})^{2}

2^{n}u^{n}v^{n}

(2^{n2}u^{n} + v^{n})^{2}

(2^{k}q^{2} – v^{n} )^{2}_{}

2 ^{k+2}q^{2}v^{n}

(2^{k}q^{2} + v^{n} )^{2}

(2^{k} – 1)^{2}

2 ^{k+2}

(2k + 1)^{2}

If 2^{n}u^{n}v^{n} =
4x^{2}y^{2} then u^{n} = 4x^{2}y^{2}
/ 2^{n}v^{n} but v^{n}
= y^{2 }so u^{n} = 4 x^{2} / 2^{n} and u =
x^{2/n}/ 2^{(n2)/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 b^{n }= 4x^{2}y^{2}and resolving
the question as to how the RHS can have a rational nth rootwhen n > 2. The proof divides 4x^{2}y^{2}^{ } into odd and even factors: 4x^{2}y^{2}=4p^{2}q^{2},
where p2 only and q∤ 2
and where p^{2}=
2^{k}, producing the
expression 4x^{2}y^{2}= 2^{k+2}q^{2}y^{2}.
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 = log_{2}b – log_{2}s –
log_{2}v. 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, a^{n }= (2^{k} – 1)^{2}
and c^{n} = (2^{k} + 1)^{2}.
The remainder of the proof shows that (2^{k}
– 1)^{2}and (2^{k} +
1)^{2} cannot both have rational nth roots. The consequence is that
the condition a, b, c,
n Î ℕ_{1} cannot be true and FLT
must be true. In proof 2, FLT is redefined as NKSabcn a^{n}
+b^{n} = c^{n }Qc. The corollary
is that CQcNSabcn a^{n} +b^{n}
= c^{n }. 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
T_{1}: SABC A^{2 }+ B^{2} = C^{2}
A,B,C Î ℕ_{1} A,B,C HNCF
T_{2}: Pxy (x^{2}  y^{2})^{2}
+ 4x^{2}y^{2} = (x^{2} + y^{2})^{2} x,y Î ℕ_{1} x,y HNCF, x2, y ∤
2
Q: Sabcn a^{n} + b^{n} = c^{n}
Qc: n
> 2 a, b,
c, n Î ℕ_{1} a,b,c HNCF
Lemma 1: C n ∤ 2 x ∉ ℚ
Line

Step

Justification

1

x^{2}
= p^{2}q^{2}

df in. d1

2

p^{2}=
2^{k}

df in. d6

3

x^{2}
= 2^{k}q^{2}

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 = 2^{k/2}q

3 square root

8

C n ∤ 2
x ∉
ℚ

6,7

Proof 1: C n ∤ 2 FLT
Line

Step

Justification

1

CK n ∤ 2 P_{1} x ∉
ℚ

Lemma 1

2

K n ∤ 2 P_{1}^{o}

Supposition

3

x ∉ ℚ

1.2 MP C exp.

4

T_{2}

Theorem in.

5

CT_{2}c_{2}_{}

T_{2}
condition set

6

c_{2}

4,5 MP Cexp.

7

Cc_{2}
x Î ℕ_{1}

c_{2}
condition on x

8

x Î ℕ_{1}

6,7 MP C exp.

9

x Î ℚ

8 Cℕ_{1}ℚ

10

x Î ℚ

9 tr.

11

K x Î ℚ x ∉
ℚ

3,10 K in.

12

C K n ∤
2 P_{1} K x Î ℚ x ∉
ℚ

3.11 C in.

13

N K n ∤
2 P_{1}

12 ENpCpKqNq

14

C n ∤
2 NP_{1}

13 De Morgan

15

P_{1 }=
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

x^{2 }=
2^{k}q^{2 }

Lemma 1

2

2^{k}
= x^{2}/q^{2}

rearrange

3

log_{2}2^{k}
= log_{2}(x^{2}/q^{2})

Take logs base 2

4

k = 2log_{2}x
– 2log_{2}q

Evaluate logs

5

CK q > 1 q
∤
2 log_{2}q ∉ ℚ

Log of odd number

6

q ∤
2

C6 cond. On q

7

C q > 1 log_{2}q
∉
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 2log_{2}x
∉
ℚ

Log of irrational #

12

2log_{2}x ∉
ℚ

10, 11 C exp.

13

C n ∤ 2
2log_{2}x ∉ ℚ

9, 12 C in.

14

C
2log_{2}x ∉
ℚ k ∉
ℚ

4

15

C 2log_{2}x
∉
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 =
log_{2}b – log_{2}s – log_{2}v
Line

Step

Justification

1

b = 2uv

d7 df in.

2

u = rs

d4 df in.

3

u^{n}
= 4x^{2}/2^{n}

d1, d2 dfs in.

4

x^{2}
= 2^{k}q^{2}

Lemma 1

5

u^{n}
=2^{k+2} q^{2}/2^{n}

3, 4

6

r^{n}s^{n}
= 2^{k+2}
q^{2}/2^{n}

2, 5

7

2^{n}r^{n}/2
^{k+2 } = q^{2}/s^{n}

6 parity error

8

2^{n}r^{n}
= 2^{k+2}

7 PE resolved

9

q^{2}
= s^{n}

7 PE resolved

10

r^{n}
= 2^{k+2 }^{n}

8 rearrange

11

k = mn  2

D5 df in.

12

r^{n}
=2^{mn} ^{ n}

10, 11

13

r = 2^{m1}

12 nth root

14

2r = 2^{m}

13

15

b = 2rsv

1, 2

16

b = 2^{m}sv

14, 15

17

2^{m}
= b/sv

16 rearrange

18

log_{2}2^{m}
= log_{2}b – log_{2}s – log_{2}v

17 take logs base 2

19

m = log_{2}b
– log_{2}s – log_{2}v

evaluate

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

Step

Justification


1

m = log_{2}b
– log_{2}s – log_{2}v

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

EA^{2}
b ≠ 2^{l} s ≠ 1 v
≠ 1 m ∉
ℚ

1 rationality of logs


6

EA^{2}
b ≠ 2^{l} s ≠ 1 v
≠ 1 n ∉
ℚ

4, 5 E subs.


7

K^{2} b = 2^{l} s = 1 v = 1 ^{o}

S_{1}

Supposition

8

x^{2} = 2^{k}q^{2}

Lemma 1


9

q^{2} = s^{n}

Lemma 3


10

q^{2} = 1

7 supposition


11

x^{2} + y^{2} = 2^{k
}+1

7, 8, 10


12

x^{2
} y^{2} = 2^{k } 1

7, 8, 10


13

(x^{2} + y^{2 })^{2}=
c^{n }

d1, d2 defs in.


14

(x^{2} – y^{2})^{2 }=
a^{n}

d1, d2 defs in


15

c = (2^{k }+1)^{2/n}

13 nth root


16

a = (2^{k } 1)^{2/n}

14 nth root


17

K c = (2^{k }+1)^{2/n}
a = (2^{k } 1)^{2/n}

15, 16 K in.


18

CS_{1} K c = (2^{k }+1)^{2/n} a =
(2^{k } 1)^{2/n}

7, 17 C in.


19

ES_{1} n Î ℚ

6, 7


20

C n Î ℚ S_{1}

19 CEpqCpq


21

C n Î ℚ K c = (2^{k }+1)^{2/n} a = (2^{k } 1)^{2/n}

18, 20 trans.

Lemma 5: NSfgK f^{n}
– g^{n} = 2 K f Î
ℕ_{1
}g
Î ℕ_{1}
Line

Step

Justification


1

SfgKf^{n} –g^{n}
= 2 Kf Î ℕ_{1 }g Î ℕ_{1}^{o}

P_{1}


2

fgKf^{n} –g^{n} = 2 Kf Î ℕ_{1 }g Î ℕ

P_{2}

1 S exp.

3

f^{n} –g^{n} = 2

P_{3}

2 K exp.

4

f^{n} = 2^{k} + 1

P_{4}

d9 df in.

5

g^{n} = 2^{k}  1

P_{5}

d9 df in.

6

f ∤ 2

P_{6}

4 RHS odd

7

g ∤ 2

P_{7}

5 RHS odd

8

n > 2

P_{8}

c2 in.

9

K^{2} f ∤ 2
g ∤ 2
n > 2

P_{9}

6, 7, 8 K^{2} in.

10

C P_{9} f^{n} –g^{n}
> 2

P_{10}

9 f >g. g > 1

11

f^{n} –g^{n} > 2

P_{11}

9, 10 C exp.

12

f^{n}
–g^{n} ≠ 2

P_{12}

11

13

K f^{n} –g^{n} = 2 f^{n}
–g^{n} ≠ 2

P_{13}

3, 12 K in.

14

Sfg K f^{n}
–g^{n} = 2 f^{n} –g^{n} ≠ 2

P_{14}

1. S in.

15

C P_{1}K
P_{12}N P_{12}

P_{15}

14

16

N P_{1}

P_{16}

15 C exp.

17

NSfgK f^{n}
–g^{n} = 2 K f Î ℕ_{1 }g Î ℕ

P_{17}

16 expand

Lemma 6: A(2^{k}
+ 1)^{1/n} ∉
ℚ
(2^{k }1)^{1/n} ∉ ℚ
Line

Step

Justification


1

K(2^{k}
+1)^{1/n }Î ℚ
(2k – 1)^{1/n} Î ℚ
^{o}

P_{1}

Supposition

2

(2^{k}
+1)^{1/n }∉ ℚ = SfK 2^{k}
+1= f^{n} fÎℕ_{1 }

P_{2}

d10 df in.

3

(2^{k}
 1)^{1/n }∉ ℚ = SfK 2^{k}
 1= g^{n} fÎℕ_{1 }

P_{3}

d10 df in.

4

(2^{k} +1)^{1/n }∉
ℚ

P_{4}

1 K exp.

5

(2k – 1)^{1/n} ∉ ℚ

P_{5}

1 K exp.

6

SfK 2^{k} +1= f^{n} fÎℕ_{1 }

P_{6}

2, 4 df sub.

7

SgK 2^{k}  1= g^{n} fÎℕ_{1 }

P_{7}

3.5 df sub

8

f  K
2^{k} +1= f^{n} fÎℕ_{1 }

P_{8}

6 S exp.

9

g K 2^{k}  1= g^{n} gÎℕ_{1 }

P_{9}

7 S exp.

10

2^{k} +1= f^{n}

P_{10}

8 K exp.

11

2^{k}  1= g^{n}

P_{11}

9 K exp

12

f^{n} – g^{n} = 2

P_{12}

10, 11 diff

13

fÎℕ_{1 }

P_{13}

8 K exp.

14

gÎℕ_{1 }

P_{14}

9 K exp.

15

K fÎℕ_{1 }gÎℕ_{1 }

P_{15}

13, 14 K in.

16

K f^{n} – g^{n} = 2K fÎℕ_{1 }gÎℕ_{1 }

P_{16}

12, 15 K in.

17

SfgK f^{n} – g^{n} = 2K
fÎℕ_{1 }gÎℕ_{1 }

P_{17}

16 S in.

18

CP_{1}
SfgK f^{n} – g^{n} = 2K fÎℕ_{1 }gÎℕ_{1 }

P_{18}

1, 17 C in.

19

NSfgK f^{n}
– g^{n} = 2K fÎℕ_{1 }gÎℕ_{1 }

P_{19}

Lemma 5

20

NP_{1}

P_{20}

18, 19 C exp.

21

NK(2^{k}
+1)^{1/n }Î ℚ
(2k – 1)^{1/n} Î ℚ

P_{21}

20 NP_{1 }expand

22

A(2^{k}
+1)^{1/n }∉ ℚ
(2k – 1)^{1/n} ∉ ℚ

P_{22}

21 De Morgan

Lemma 7: E w^{1/n}
∉
ℚ
w^{2/n }∉ ℚ
Line

Step

Justification

1

n > 2 ^{o}

Supposition

2

C w^{1/n }∉
ℚ w^{2/n }∉ ℚ

1 2/n < 1

3

C w^{2/n }∉ ℚ w^{1/n }∉
ℚ

1 1/n < 1

4

E w^{1/n }∉
ℚ w^{2/n }∉ ℚ

2, 3 E in.

5

C n > 2 E
w^{1/n }∉ ℚ w^{2/n }∉ ℚ

1, 4 C in.

6

n > 2

C_{2 }condition

7

E w^{1/n }∉
ℚ w^{2/n }∉ ℚ

5, 6 C exp.

Lemma 8: C nÎ
ℚ A c ∉
ℚ a ∉ ℚ
Line

Step

Justification

1

E (2^{k}
+1)^{1/n }∉ ℚ (2k
– 1) ^{2/n} ∉ ℚ

Lemma 7 subs.

2

A (2^{k}
+1) ^{2/n }∉ ℚ (2k
– 1) ^{2/n} ∉ ℚ

Lemmas 6, 7

3

n
Î ℚ ^{o}

Suppostion

4

K c = (2^{k} +1) ^{2/n } a = (2k – 1) ^{2/n}

Lemma 4

5

c
= (2^{k} +1) ^{2/n }

4 K exp.

6

a = (2k – 1) ^{2/n}

4 K exp.

7

A (2^{k} +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 (2^{k}
+1)^{1/n }∉ ℚ (2k
– 1) ^{2/n} ∉ ℚ

Lemma 7 subs.

2

A (2^{k}
+1) ^{2/n }∉ ℚ (2k
– 1) ^{2/n} ∉ ℚ

Lemmas 6, 7

3

n
Î ℚ ^{o}

Suppostion

4

K c = (2^{k} +1) ^{2/n } a = (2k – 1) ^{2/n}

Lemma 4

5

c
= (2^{k} +1) ^{2/n }

4 K exp.

6

a = (2k – 1) ^{2/n}

4 K exp.

7

A (2^{k} +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 a^{n} +b^{n} = c^{n}
K^{2 }n > 2 a, b, c, n Î
ℕ_{1
}a, b, c HNCF
Line

Step

Justification


1

Q = df KSabcn a^{n} +b^{n} = c^{n }Qc

P_{1}

Q df in.

2

Qc = df K^{2 }n > 2 a, b, c, n Î ℕ_{1 }a, b, c HNCF

P_{2}

Qc df in

3

Q ^{o}

P_{3}

Supposition

4

Qc

P_{4}

1 K exp.

5

K^{3} a Îℕ_{1 }b Îℕ_{1 }c Îℕ_{1 }n Îℕ_{1 }

P_{5}

2, 4 expand K
exp.

6

C n Î ℚ A c ∉ ℚ a ∉
ℚ

P_{6}

Lemma 8

7

n Î ℕ_{1}

P_{7}

5 K exp.

8

n ∉ ℚ

P_{8}

7 C ℕ_{1}ℚ

9

A
c ∉
ℚ a ∉ ℚ

P_{9}

6, 8 C exp.

10

K c Î ℕ_{1
}a
Î ℕ_{1
}

P_{10}

5 K exp.

11

A
c ∉
ℕ_{1 }a ∉ ℕ_{1
}

P_{11}

9 C x ∉ ℚ x ∉ ℕ_{1 }

12

N K c Î ℕ_{1
}a
Î ℕ_{1
}

P_{12}

10 De Morgan

13

K
P_{10}N P_{10}

P_{13}

10, 12 K in

14

CQ K P_{11}N
P_{11}

P_{14}

3, 13 C in.

15

NQ

P_{15}

14 ECpNpNp

16

NKSabcn a^{n}
+b^{n} = c^{n }Qc

P_{16}

15 expand

17

FLT = df NKSabcn a^{n}
+b^{n} = c^{n }Qc

P_{17}

16 df in.

18

FLT

P_{18}

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 A^{2 }+ B^{2} = C^{2} : Pythagorean form
a,
b, c, n, where a^{n} + b^{n} = c^{n} : 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: a^{n} = A^{2}, b^{n}
= B^{2}, c^{n} = C^{2}
d2: A = x^{2}
 y^{2}, B = 2xy, C = x^{2} + y^{2}
d3: x^{2} = p^{2}q^{2}
d4: u = rs
d5: k = mn – 2
d6: p^{2}= 2^{k}
d7: b = 2uv
d8: v^{n} = y^{2}
d9: f^{n} = 2^{k} + 1, g^{n} = 2^{k}
 1
Conditions
c1: A, B, C Î ℕ_{1 }, A, B, C HNCF
c2: a, b, c, Î ℕ_{1
}, a, b, c HNCF, a ∤ 2, c ∤
2, b  2
c3: n Î ℕ_{1}, n > 2_{ }
c4: x, y Î ℕ_{1 }, x, y HNCF , x  2,
y ∤ 2
c5: u, v Î ℕ_{1 }, u, v HNCF , u  2, v
∤ 2
c6: p2 only, q ∤ 2
c7: r2 only, s ∤ 2
c8: l,
m Î ℕ_{1}
c9: f, g Î ℝ