1

Keywords: Fermat’s last theorem, Pascal triangle, Pythagorean equation, Figurate polynomials

Subject classification: 11D41, 11B83, 11C08

Author: Vladimir Godovalov

Independent researcher

BS in E&E Engineering, Moscow State Technological Aviation University, 1990-1996

Email: vgodovalov@yahoo.com

Language: English

Comparative analysis of 𝒙𝟑 + 𝒚𝟑 = 𝒛𝟑 and 𝒙𝟐 + 𝒚𝟐 = 𝒛𝟐 in the Interconnected Sets

Vladimir Godovalov

Abstract. This paper introduces an innovative technique of study 𝑧3 − 𝑥3 = 𝒚𝟑 on the subject of its

insolvability in integers. Technique starts from building the interconnected, third degree sets:

𝐴3 = 𝑎𝑛 𝑎𝑛 = 𝑛3, 𝑛 ∈ ℕ , 𝐵3 = 𝑏𝑛 𝑏𝑛 = 𝑎𝑛+1 − 𝑎𝑛 , 𝐶3 = 𝑐𝑛 𝑐𝑛 = 𝑏𝑛+1 − 𝑏𝑛 and 𝑃3 = {6}

wherefrom we get 𝑎𝑛 and 𝑏𝑛 expressed as figurate polynomials of third degree, a new finding in

mathematics. This approach and the results allow us to investigate equation 𝑧3 − 𝑥3 = 𝑦 in these

interconnected sets 𝐴3 and 𝐵3, where 𝑧3 ∧ 𝑥3 ∈ 𝐴3, 𝑦 ∈ 𝐵3. Further, in conjunction with the new

Method of Ratio Comparison of Summands and Pascal’s rule, we finally prove inability of 𝑦 = 𝒚𝟑. After

we test the technique, applying the same approach to 𝑧2 − 𝑥2 = 𝑦 where we get family of primitive

𝑧2 − 𝑥2 = 𝑦2 as well as introduce conception of the basic primitiveness of 𝑧′2 − 𝑥′2 = 𝑦2 for

𝑧′ − 𝑥′ = 1 and the dependant primitiveness of 𝑧′2 − 𝑥′2 = 𝑦2 for co-prime 𝑥,𝑦, 𝑧 and 𝑧′ − 𝑥′ > 1.

Part 1.0 Third degree figurate polynomials getting

This technique requires a new terminology. So, in this paper we extended notation for the sets by

attaching a degree number to the symbols of set, introduced the term “figurate polynomial” and

designated a two-index symbol to the figurate numbers.

Build the table of elements of the sets ℕ,𝐴3,𝐵3,𝐶3,𝑃3:

Table 1

𝑛, 𝑛 ∈ ℕ

𝐴3, 𝐴3 = 𝑎𝑛 𝑎𝑛 = 𝑛3

𝐵3, 𝐵3 = 𝑏𝑛 𝑏𝑛 = 𝑎𝑛+1 − 𝑎𝑛

𝐶3,

𝐶3 = 𝑐𝑛 𝑐𝑛 = 𝑏𝑛+1 − 𝑏𝑛

𝑃3, 𝑃3 = 6

0

0 1

0 6

1

1 7

6 6

2

8 19

12 6

3 27 18

2

37 6 4

64 61

24 6

5

125 91

30 6

6

216 127

36 6

7

343 169

42 6

8

512 217

48 6

9

729 271

54 6

10

1000 331

60 6

… … … … … ∞ ∞ ∞ ∞ ∞

The first column is the set ℕ: 0, 1, 2, 3, 4,…𝑛.

Second column are the set 𝐴3 elements: 𝑎0 = 03 = 0, 𝑎1 = 13 = 1, 𝑎2 = 23 = 8, 𝑎3 = 33 = 27,

𝑎4 = 43 = 81,… 𝑎𝑛 = 𝑛3.

Subtraction of adjacent elements of the set 𝐴3 are the set 𝐵3 elements: 𝑏0 = 𝑎1 − 𝑎0 = 1, 𝑏1 = 𝑎2 −

𝑎1 = 7, 𝑏2 = 𝑎3 − 𝑎2 = 19,… 𝑏𝑛 = 𝑎𝑛+1 − 𝑎𝑛 .

Subtraction of adjacent elements of the set 𝐵3 are the set 𝐶3 elements: 𝑐0 = 𝑏1 − 𝑏0 = 6, 𝑐1 = 𝑏2 −

𝑏1 = 12,… 𝑐𝑛 = 𝑏𝑛+1 − 𝑏𝑛 .

Subtraction of adjacent elements of the set 𝐶3 are the set 𝑃3 elements: 𝑝 = 𝑐1 − 𝑐0 = 6, 𝑝 = 𝑐2 −

𝑐1 = 6,…𝑝 = 𝑐𝑛+1 − 𝑐𝑛 = 6.

Apply the algorithm of getting of 𝑎𝑛 и 𝑏𝑛 through elements of the set 𝑃3.

We have for:

the set 𝐶3: 𝑐𝑛 = 6𝑛;

the set 𝐵3: 𝑏1 = 𝑏0 + 6 ∙ 1, 𝑏2 = 𝑏0 + 6 1 + 2 , 𝑏3 = 𝑏0 + 6(1 + 2 + 3);

the set 𝐴3: 𝑎1 = 𝑎0 + 𝑏0 , 𝑎2 = 2𝑏0 + 6 ∙ 1, 𝑎3 = 3𝑏0 + 6 2 ∙ 1 + 2 , 𝑎4 = 4𝑏0 + 6(3 ∙ 1 + 2 ∙ 2 + 3).

Make analysis of the results:

- for the set 𝐵3 we have the sequence: 1, 1 + 2 , 1 + 2 + 3 , what is the triangular figurate

numbers – 1,3,6. Designate their 𝑘𝑛(2)

, where 𝑛 – sequence number, (2) – figurate row number.

Since 𝑏0 = 1, and with help of induction method, we get the common formula of the set 𝐵3

elements:

𝑏𝑛 = 1 + 6𝑘𝑛(2)

(1.0)

Triangular figurate numbers 𝑘𝑛(2)

are the numbers of third Pascal’s triangle diagonal [1] also known as

the OEIS library sequence A000217 [2].

3

- For the set 𝐴3 we have: coefficients of 𝑏0 − 1,2,3,4 which correspond to index 𝑛 of each 𝑎𝑛 ,

and sequence: 1, 2 ∙ 1 + 2 , 3 ∙ 1 + 2 ∙ 2 + 3 what is the tetrahedral figurate numbers –

1,4,10. Designate their 𝑘𝑛(3)

, and with help of induction method, get the common formula of the

set 𝐴3 elements:

𝑎𝑛 = 𝑛 + 6𝑘𝑛−1(3)

(1.1)

Tetrahedral figurate numbers 𝑘𝑛(3)

are the numbers of forth Pascal’s triangle diagonal [2] also known as

the OEIS library sequence A000292 [3].

Finally, using the table’s dependences write two formulas:

1) Subtraction of two adjacent set 𝐴3 elements

𝑎𝑛+1 − 𝑎𝑛 = 1 + 6𝑘𝑛(2)

(1.2)

is a single set 𝐵3 element 𝑏𝑛 , where 𝑛 ≥ 0.

2) Subtraction of two non-adjacent set 𝐴3 elements

𝑎𝑛+𝑘 − 𝑎𝑛 = 1 + 6𝑘𝑖(2) 𝑛+𝑘−1

𝑖=𝑛 (1.3)

is a 𝑘 − sum of the set 𝐵3 elements 𝑏𝑛 , where 𝑛 ≥ 0, 𝑘 ≥ 1.

Objectivities of this technique

a) Based on (1.2) and (1.3) we can study any 𝑧3 − 𝑥3 = 𝑦 in order to find out whether 𝑦 = 𝒚𝟑,

taking 𝑧3 ∧ 𝑥3 ∈ 𝐴3, 𝑦 ∈ 𝐵3 and 𝑦 is odd.

b) Either 𝑧3 , 𝑥3 and 𝑦 have got additional properties due to their expression in the figurate

polynomials form (1.0) and (1.1)

c) For any 𝑧3 there are involved all possible combinations of 𝑥3 and 𝑦.

d) The equation 𝑧3 − 𝑥3 = 𝑦 always has solutions in integers

These are fundamental differences between these principles and currently existing.

Part 1.1 Set B3 investigation

The obvious first step to find out whether 𝑦 = 𝒚𝟑 in 𝑧3 − 𝑥3 = 𝑦 is analysis of each single element 𝑏𝑛 of

the set 𝐵, taking 𝑧3 = 𝑎𝑛+1 , 𝑥3 = 𝑎𝑛 and 𝑦 = 𝑏𝑛 according to (1.2). So we come to the condition

𝑏𝑛 ∈ 𝐴3 ↔ 𝑧3 − 𝑥3 = 𝒚𝟑 (1.4)

Since the condition 𝑏𝑛 ∈ 𝐴3 means equality of some 𝑏𝑛 to some 𝑎𝑛 then we should differentiate

indexes 𝑛 and rewrite (1.4)

𝑏𝑛 = 𝑎𝑛 ′ ↔ 𝑧3 − 𝑥3 = 𝒚𝟑 (1.5)

4

To find solution (1.5) we need transform the general condition 𝑏𝑛 = 𝑎𝑛 ′ into particular case of some

common form, so express these elements in the kind of a set 𝐵3 figurate polynomial, thus

𝑏𝑛 = 1 + 6𝑘𝑛(2)

and

𝑎𝑛 ′ = 𝑛′ + 6𝑘𝑛 ′−1

(3)

and finally the required form

𝑎𝑛 ′ = 1 + 6 𝑘𝑛 ′−1

(3)+𝑛′ − 1

6

Rewrite (2.2) in kind of

𝑘𝑛(2)

= 𝑘𝑛 ′−1

(3)+𝑛′ − 1

6 ↔ 𝑏𝑛 = 𝑎𝑛 ′ (1.6)

So we have the task to solve: can we get a triangular figurate number 𝑘𝑛(2)

from the sum of a tetrahedral

figurate number 𝑘𝑛 ′−1

(3) and

𝑛 ′−1

6 ?

Part 1.2 Searching 𝒃𝒏 = 𝒂𝒏′ in the 𝑨𝟑 and 𝑩𝟑 sets

According to (1.6) we should find some 𝑛 and 𝑛′ if

𝑘𝑛(2)

= 𝑘𝑛 ′−1

(3)+𝑛′ − 1

6 (1.7)

is true or prove that such are absent.

To solve this problem we use the Method of Ratio Comparison of Summands and Pascal’s rule, a one

new tool. The left part of (1.7) is the sequence of triangular figurate numbers 𝑘𝑛(2)

presented by the

third diagonal of Pascal triangle, where the formula to get these (each number is a sum of two above it)

is 𝑛+12 = 𝑛 +

𝑛(𝑛−1)

2. Rewrite 𝑛 +

𝑛(𝑛−1)

2 into similar specific form of kind “number+coefficient ∙

number” but with the integer coefficient, get: 𝑛 +𝑛(𝑛−1)

2=

𝑛

2+ 𝑛 ∙

𝑛

2 , and write (1.7) as

𝑛

2+ 𝑛 ∙

𝑛

2= 𝑘

𝑛 ′−1

(3)+

𝑛 ′−1

6 (1.8)

Disclosing the figurate number 𝑘𝑛 ′−1

3 =𝑛 ′ 𝑛 ′ 2−1

6 , and also rewriting right part (1.8) into the specified

form, get: 𝑘𝑛 ′−1

3 +𝑛 ′−1

6=

𝑛 ′−1

6+ 𝑛′ 𝑛′ + 1 ∙

𝑛 ′−1

6, write (1.8) as

𝑛

2+ 𝑛 ∙

𝑛

2=

𝑛 ′−1

6+ 𝑛′ 𝑛′ + 1 ∙

𝑛 ′−1

6 (1.9)

As it is seen, firstly, both sides (1.9) are functions of one variable in kind of sum of two summands and

secondly each function generates particular pair of numbers for summing. Use Method of Ratio

Comparison of Summands, which divides the greater summand on the smaller in each side (1.9), then

5

compares these ratios and with help of Lemma1 provides a strong conclusion. So, introducing the ratio

for left side (1.9) as 𝑅3, get

𝑅3 = 𝑛

For the right side (1.9) accordingly get

𝑅3′ = 𝑛′(𝑛′ + 1)

Lemma 1:

Values of functions 𝑛 ′−1

6+ 𝑛′ 𝑛′ + 1 ∙

𝑛 ′−1

6 ≠

𝑛

2+ 𝑛 ∙

𝑛

2 for any 𝑛′ ∧ 𝑛 ∈ ℕ, if 𝑅3′ ≠ 𝑅3 for

common 𝑛.

Proof:

Let for some 𝑛′ value of function 𝑛 ′−1

6+ 𝑛′ 𝑛′ + 1 ∙

𝑛 ′−1

6 =

𝑛

2+ 𝑛 ∙

𝑛

2 for another 𝑛, where 𝑛 > 𝑛′ .

Then using equipollent changing of each summand on some value 𝑗, the function

𝑛 ′−1

6+ 𝑗 + 𝑛′ 𝑛′ + 1 ∙

𝑛 ′−1

6− 𝑗 eventually takes form

𝑛 ′

2+ 𝑛′ ∙

𝑛 ′

2 where 𝑛′ = 𝑛. Obviously here,

the ratio 𝑅3′ = 𝑛′ , and since the values of these functions are equal , hence, for this 𝑛′ the ratio of

𝑛 ′−1

6+ 𝑛′ 𝑛′ + 1 ∙

𝑛 ′−1

6 will be the same. However it is known that for all 𝑛′ the ratio of this function

is 𝑅3′ = 𝑛′(𝑛′ + 1). But the set ℕ does not have such 𝑛′ for which will exist simultaneously two various

ratios of the same function. From here follows nonexistence of such 𝑛′ .

Lemma 1 is proved.

So, we never get a triangular figurate number 𝑘𝑛(2)

from the sum of a tetrahedral figurate number 𝑘𝑛 ′−1

(3)

and 𝑛 ′−1

6.

Conclusion: 𝑏𝑛 ≠ 𝑎𝑛 ′ ↔ 𝑧3 − 𝑥3 ≠ 𝒚𝟑

The set 𝐵3 does not have even a single 𝑏𝑛 equal to 𝑎𝑛 ′ of the set 𝐴3. It means that 𝑦 ≠ 𝒚𝟑, hence,

𝑧3 − 𝑥3 = 𝒚𝟑 does not exist in the sets 𝐴3 and 𝐵3 for 𝑧 − 𝑥 = 1.

Part 2.0 Second degree figurate polynomials getting

Test our technique by applying the same principles and methods to 𝑧2 − 𝑥2 = 𝑦 in the interconnected

sets 𝐴2 and 𝐵2, where 𝑧2 ∧ 𝑥2 ∈ 𝐴2, 𝑦 ∈ 𝐵2 and 𝑦 is odd.

Build the table of elements of the sets ℕ,𝐴2,𝐵2,𝑃2:

Table 2

𝑛, 𝑛 ∈ 𝑁 𝐴2 = 𝑎𝑛 𝑎𝑛 = 𝑛2

A2,

𝐵2 = 𝑏𝑛 𝑏𝑛 = 𝑎𝑛+1 − 𝑎𝑛

𝐵2,

𝑃2 = 2

𝑃2,

0

0 1

2

1

1 3

2

6

The first column is the set ℕ: 0, 1, 2, 3, 4,…𝑛.

Second column are the set 𝐴2 elements: 𝑎0 = 02 = 0, 𝑎1 = 12 = 1, 𝑎2 = 22 = 4, 𝑎3 = 32 = 9, 𝑎4 =

42 = 16,… 𝑎𝑛 = 𝑛2.

Subtraction of adjacent elements of the set 𝐴2 are the set 𝐵2 elements: 𝑏0 = 𝑎1 − 𝑎0 = 1, 𝑏1 = 𝑎2 −

𝑎1 = 3, 𝑏2 = 𝑎3 − 𝑎2 = 5,… 𝑏𝑛 = 𝑎𝑛+1 − 𝑎𝑛 .

Subtraction of adjacent elements of the set 𝐵2 are the set 𝑃2 elements: 𝑝 = 𝑏1 − 𝑏0 = 2, 𝑝 = 𝑏2 −

𝑏1 = 2,…𝑝 = 𝑏𝑛+1 − 𝑏𝑛 = 2.

Apply the algorithm of getting of 𝑎𝑛 и 𝑏𝑛 through elements of the set 𝑃2.

We have for:

the set 𝐵2: 𝑏𝑛 = 2 𝑛 +1

2 ;

the set 𝐴2: 𝑎1 = 𝑎0 + 𝑏0 , 𝑎2 = 𝑎0 + 𝑏0 + 2 1 + 1 ∙1

2 , 𝑎3 = 𝑎0 + 𝑏0 + 2 1 + 2 + 2 ∙

1

2 , 𝑎4 = 𝑎0 +

𝑏0 + 2 1 + 2 + 3 + 3 ∙1

2 .

Make analysis of the results:

- for each set 𝐵2 element we have

𝑏𝑛 = 1 + 2𝑛 (2.0)

- for the elements 𝑎1 ,𝑎2 ,𝑎3 ,𝑎4 we have: 𝑎0 = 0, 𝑏0 = 1, coefficients of 1

2− 1,2,3 which

correspond to index of each 𝑎𝑛 as 𝑛 − 1, and the sequence: 1, 1 + 2 , 1 + 2 + 3 what are

the triangular figurate numbers – 1,3,6. Designate their as 𝑘𝑛(2)

, and with help of induction

method, get the common formula of the set 𝐴2 elements:

2

4 5

2

3

9 7

2

4

16 9

2

5

25 11

2

6

36 13

2

7

49 15

2

8

64 17

2

9

81 19

2

10

100 21

2

… … … … ∞ ∞ ∞ ∞

7

𝑎𝑛 = 1 + 2 𝑘𝑛−1(2)

+𝑛 − 1

2

or

𝑎𝑛 = 𝑛 + 2𝑘𝑛−1(2)

(2.1)

Triangular figurate numbers 𝑘𝑛(2)

are the numbers of third Pascal’s triangle diagonal [1] also known as

the OEIS library sequence A000217 [2].

Finally, using the table’s dependences write two formulas:

1) Subtraction of two adjacent set 𝐴2 elements

𝑎𝑛+1 − 𝑎𝑛 = 1 + 2𝑛 (2.2)

is a single set 𝐵2 element 𝑏𝑛 , where 𝑛 ≥ 0.

2) Subtraction of two non-adjacent set 𝐴2 elements

𝑎𝑛+𝑘 − 𝑎𝑛 = 1 + 2𝑗 𝑛+𝑘−1𝑖=𝑛 (2.3)

is a 𝑘 − sum of the set 𝐵2 elements 𝑏𝑛 , where 𝑛 ≥ 0, 𝑘 ≥ 1.

Part 2.1 Set B2 investigation

The obvious first step to find out whether 𝑦 = 𝒚𝟐 in 𝑧2 − 𝑥2 = 𝑦 is analysis of each single element 𝑏𝑛 of

the set 𝐵2, taking 𝑧2 = 𝑎𝑛+1 , 𝑥2 = 𝑎𝑛 and 𝑦 = 𝑏𝑛 according to (2.2). So we come to the condition

𝑏𝑛 ∈ 𝐴2 ↔ 𝑧2 − 𝑥2 = 𝒚𝟐 (2.4)

Since the condition 𝑏𝑛 ∈ 𝐴2 means equality of some 𝑏𝑛 to some 𝑎𝑛 then we should differentiate

indexes 𝑛 and rewrite (2.4)

𝑏𝑛 = 𝑎𝑛 ′ ↔ 𝑧2 − 𝑥2 = 𝒚𝟐 (2.5)

To find solution (2.5) we need transform the general condition 𝑏𝑛 = 𝑎𝑛 ′ into particular case of some

common form, so express these elements in the kind of a set 𝐵2 figurate polynomial, thus

𝑏𝑛 = 1 + 2𝑛

and

𝑎𝑛 ′ = 𝑛′ + 2𝑘𝑛 ′−1

(2)

where finally the required form is

𝑎𝑛 ′ = 1 + 2 𝑘𝑛 ′−1

(2)+𝑛′ − 1

2

Rewrite (2.2) in kind of

8

𝑛 = 𝑘𝑛 ′−1

(2)+𝑛′ − 1

2 ↔ 𝑏𝑛 = 𝑎𝑛 ′ (2.6)

So we have got the task to solve: can we get an integer 𝑛 from the sum of a triangular figurate number

𝑘𝑛 ′−1

(2) and

𝑛 ′−1

2 ?

Part 2.2 Searching 𝒃𝒏 = 𝒂𝒏′ in the 𝑨𝟐 and 𝑩𝟐 sets

According to (2.6) we should find some 𝑛 and 𝑛′ if

𝑛 = 𝑘𝑛 ′−1

(2)+𝑛′ − 1

2 (2.7)

is true or prove that such are absent.

Find the ratio 𝑅2 for left part (2.7). Since the sequence of integers 𝑛 is the second diagonal of Pascal

triangle with formula 𝑛+11 = 1 + 𝑛, then 𝑅2 =

𝑛

1= 𝑛.

Unveiling 𝑘𝑛 ′−1

(2)=

𝑛 ′ (𝑛 ′−1)

2, rewrite the right part (2.7) into the specified form,

get: 𝑘𝑛 ′−1

(2)+

𝑛 ′−1

2=

𝑛 ′−1

2+ 𝑛′ ∙

𝑛 ′−1

2, wherefrom find the ratio: 𝑅2′ = 𝑛′ . Since the ratios 𝑅2 = 𝑅2′ for

common 𝑛, then follows, there are such 𝑛 and 𝑛′ when (2.7) is true as says Lemma 1 in interpretation

for these functions.

There is no need to formulate and prove the lemma for this case, because if write (2.7) as

𝑛 =𝑛 ′ 2−1

2 (2.8)

becomes clear, that in order to get in the left side of (2.8) integer 𝑛 is enough to insert in the right side

an odd 𝑛′ . Actually, for

𝑛′ = 3: 𝑛 =32 − 1

2= 4

𝑛′ = 5: 𝑛 =52 − 1

2= 12

𝑛′ = 7: 𝑛 =72 − 1

2= 24

𝑛′ = 9: 𝑛 =92 − 1

2= 40

From here, the first four equalities of 𝑏𝑛 = 𝑎𝑛 ′ are:

𝑏4 = 𝑎3′

𝑏12 = 𝑎5′

𝑏24 = 𝑎7′

9

𝑏40 = 𝑎9′

In connection with this, write down the first four primitive Pythagorean equations, where according to

formula (2.2), get:

𝑎5′ − 𝑎4′ = 𝑏4 or in the numerical expression: 52 − 42 = 32

𝑎13′ − 𝑎12′ = 𝑏12 or in the numerical expression: 132 − 122 = 52

𝑎25′ − 𝑎24′ = 𝑏24 or in the numerical expression: 252 − 242 = 72

𝑎41′ − 𝑎40′ = 𝑏40 or in the numerical expression: 412 − 402 = 92

This technique allows us to find out and write down the sequence of Pythagorean primitive triples.

Part 3. Basic and dependant primitiveness

As it is known, if each member of a Pythagorean triple has no common factor then this equation calls

primitive. Farther we show that it is more complicated then were thought.

In the Part 1 and Part 2 we studied 𝑧3 − 𝑥3 = 𝑦 and 𝑧2 − 𝑥2 = 𝑦 for 𝑧 − 𝑥 = 1. Thus we have missed

the cases 𝑧 − 𝑥 > 1 which can have as well a number of solutions. Consider them.

Let 𝑧′3 − 𝑥′3 = 𝒚𝟑 , where 𝑧′ − 𝑥′ > 1 and 𝒚 is odd, (3.1)

then according to the factoring formula

𝑧′3 − 𝑥′3 = 𝑧′ − 𝑥′ 𝑧′2 + 2𝑧′𝑥′ + 𝑥′2

we can present the first factor as: 𝑧′ − 𝑥′ = 𝑙3 and the second as 𝑧′2 + 2𝑧′𝑥′ + 𝑥′2 = 𝑚3, from

here

𝒚𝟑 = 𝑙3 ∙ 𝑚3 (3.2)

However the result won’t change if we represent (3.2) as

𝒚𝟑 = 1 ∙ 𝑙𝑚 3

Since the first factor is 1, hence there is some 𝑧3 and 𝑥3 with 𝑧 − 𝑥 = 1 when

𝑧3 − 𝑥3 = 1 ∙ 𝒚𝟑 (3.3)

But it is false, because on condition 𝑧3 ∧ 𝑥3 ∈ 𝐴3 with 𝑧 − 𝑥 = 1 we have 𝑧3 − 𝑥3 ≠ 𝒚𝟑 as was proved

in Part 1.2

Otherwise, if we can’t get the second factor of (3.3) as 𝒚𝟑 = 𝑙3 ∙ 𝑚3 hence we have nothing to factorize

and get (3.1).

Conclusion: if 𝑧3 − 𝑥3 ≠ 𝒚𝟑 for 𝑧 − 𝑥 = 1 then 𝑧′3 − 𝑥′3 ≠ 𝒚𝟑 for 𝑧′ − 𝑥′ > 1

Now consider 𝑧′2 − 𝑥 ′2 = 𝒚𝟐, where 𝑧′ − 𝑥′ > 1 and 𝒚 is odd, (3.4)

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then according to the factoring formula

𝑧′2 − 𝑥′2 = 𝑧′ − 𝑥′ 𝑧′ + 𝑥′

we can present the first factor as: 𝑧′ − 𝑥′ = 𝑙2 and the second as 𝑧′ + 𝑥′ = 𝑚2, from here

𝒚𝟐 = 𝑙2 ∙ 𝑚2 (3.5)

However the result won’t change if we represent (3.5) as

𝒚𝟐 = 1 ∙ 𝑙𝑚 2

Since the first factor is 1, hence there are some 𝑧2 and 𝑥2 with 𝑧 − 𝑥 = 1 when

𝑧2 − 𝑥2 = 1 ∙ 𝒚𝟐 (3.6)

Actually, for any odd 𝒚𝟐 there always are such 𝑧2 ∧ 𝑥2 ∈ 𝐴2 with 𝑧 − 𝑥 = 1 when this value will be

found according (2.2).

Otherwise, if we can get the second factor of (3.6) as 𝒚𝟐 = 𝑙2 ∙ 𝑚2 hence we have nothing to factorize

and get (3.4).

Conclusion: if 𝑧2 − 𝑥2 = 𝒚𝟐 for 𝑧 − 𝑥 = 1 then 𝑧′2 − 𝑥′2 = 𝒚𝟐 for 𝑧′ − 𝑥′ > 1

Examples

Example1. Take 532 − 282 = 452,

here 532 − 282 = 53 − 28 53 + 28 , where 𝑙2 = 25 = 52 and 𝑚2 = 81 = 92

Accept the first factor is 1 then the second factor is 𝑙𝑚 2 = 2025 and there are some 𝑧2 and 𝑥2 with

𝑧 − 𝑥 = 1 when 𝑧2 − 𝑥2 = 1 ∙ 2025 = 452, to find these we need formula (2.2)

𝑎𝑛+1 − 𝑎𝑛 = 1 + 2𝑛

from it we get

𝑎𝑛+1 − 𝑎𝑛 = 1 + 2 ∙ 1012

Hence our 𝑧 and 𝑥 are

𝑧 = 𝑎𝑛+1 = 1013 and 𝑥 = 𝑎𝑛 = 1012

Now assume that 10132 − 10122 ≠ 1 ∙ 2025 ≠ 452 then we cannot factorize 1 ∙ 2025 into 25 ∙ 81 and

get 53 − 28 ∙ 53 + 28 , hence also 532 − 282 ≠ 25 ∙ 81 ≠ 452. From here follow that 10132 −

10122 = 452 is the basic primitive Pythagorean triple and 532 − 282 = 452 is its dependant primitive

Pythagorean triple, and nonexistence of the first leads to nonexistence of the second.

Obviously, the equations 𝑧2 − 𝑥2 = 𝒚𝟐 for 𝒚𝟐 = 𝑙 ∙ 𝑙 2 or 𝒚𝟐 = 𝑙2, where 𝑙 - prime number won’t

generate the dependant primitive Pythagorean triple and will exist in the single form. For example

12012 − 12002 = 492 where 492 = 7 ∙ 7 2 and 612 − 602 = 112.

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Introduce the clear definition of primitiveness of Pythagorean triples: if a Pythagorean triple has all the

members co-prime where the bases subtraction result of any two is 1 than there is the basic

primitiveness otherwise it is the dependant primitiveness.

In the end of this part give two methods of getting the primitive Pythagorean triples.

The first is for getting the basic primitive Pythagorean triples based on relation (2.8) described in Part

2.2.

The second is for getting the dependant primitive Pythagorean triples, based on (3.5). Consider it

detailed.

Let 𝑧2 − 𝑥2 = 𝑦2, where 𝑦2 = 𝑎2 ∙ 𝑏2. Take two co-prime odd numbers 𝑎 and 𝑏. Then find the centers

and offset values: 𝑑1 =𝑏2−1

2, 𝑑2 =

𝑏2+1

2 and 𝑗 =

𝑎2−1

2. And finally get: 𝑧 = 𝑑2 + 𝑗 and 𝑥 = 𝑑1 − 𝑗.

Example2. Take 𝑎 = 15 and 𝑏 = 17, then 𝑑1 =172−1

2= 144, 𝑑2 =

172+1

2= 145 and 𝑗 =

152−1

2= 112.

Finally: 𝑧 = 145 + 112 = 257 and 𝑥 = 144 − 112 = 32.

Our triple is

2572 − 322 = 2552

This equation has no signs of the basic primitiveness, find out on which it depends. Since 2552 is a set

𝐵2 element 𝑏𝑛 , find which namely it is

𝑏𝑛 = 1 + 2𝑛 = 2552 = 65025

from here get

𝑏35512 = 1 + 2 ∙ 35512

Applying (2.2) we get

355132 − 355122 = 2552

Example3. Take 𝑎 = 7 and 𝑏 = 13, then 𝑑1 =132−1

2= 84, 𝑑2 =

132+1

2= 85 and 𝑗 =

72−1

2= 24. Farther

𝑧 = 85 + 24 = 109 and 𝑥 = 84 − 24 = 60.

And finally our equitation is

1092 − 602 = 912

This equation has no signs of the basic primitiveness, find out on which it depends. Since 912 is a set 𝐵2

element 𝑏𝑛 , find which namely it is

𝑏𝑛 = 1 + 2𝑛 = 912 = 8281

from here get

𝑏4140 = 1 + 2 ∙ 4140

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Applying (2.2) we get

41412 − 41402 = 912

4.0 Conclusion

The main merit of this work is creation of the ordered structure in study those equations. It allowed us

to better understand their nature and get new results. We hope we succeeded. At least the results say

it.

References:

[1] http://en.wikipedia.org/wiki/Pascal%27s_triangle

[2] http://oeis.org/A000217

[3] http://oeis.org/A000292