Methodological Strategy for Obtaining the Possible Pythagoreans Trios and Proof Through a Criterion Applied to the Primitive Pythagorean Trios

8/2/2019 Methodological Strategy for Obtaining the Possible Pythagoreans Trios and Proof Through a Criterion Applied to the

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Methodological Strategy for Obtaining the Possible Pythagoreans Trios and Proof

through a Criterion applied to the Primitive Pythagorean Trios

Rodolfo A. Nieves Rivas

[email protected]

This article presents a method for obtaining the terms belonging to each and every one of

the possible Pythagoreans trios, which are generated by two functional formulas or

equations according to their parity (Even or odd). Then we do an analysis of the results.

And we conclude with an approach that ensures and establishes the necessary and sufficient

conditions for the characterization of all the terms of any primitive Pythagorean trios and its

application to the twin primes and Golbachs conjecture.

Keywords: methods, criterion, primitive Pythagorean trios, conjectures.
mailto:[email protected]:[email protected]:[email protected]


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INTRODUCTION

To address this issue is necessary to consider some results that have been obtained and are used by

many authors (Rada, 1992) (Joyce D.2005) because of its simplicity, effectiveness and elegance.

Such as:

all positive primitive solutions of the equation: a

2

+ b

2

= c

2

, can be expressed as follow: a = m

2

- n

2

,b = 2.mn c = m2 + n2. Where: n and m are arbitrary integers that meet three conditions:

To: m> n> 0

When (m, n) = 1

Where: m and n have different parity

The solution found by Pythagoras (Espasa, 1998) is as follows: If n is any integer number, the three

terms of trios are obtained as follows: The first term is twice the given number plus one:

a = 2.n + 1

The second term is twice the square of the given number plus its double:

b = + 2.n2

+2.n

And the third term is equal to the second term plus one:

c=2.n2 + 2.n + 1

For the development of this methodological strategy is needed a systematic presentation that

establishes a connection within the descriptive analysis in terms of obtaining the goal which has the

following order:

1.- Presentation of the method for obtaining all Pythagoreans trios.

2.- Presentation of the generating functional equations or formulas for all terms according to theparity of any possible Pythagorean triple.

3.- Presentation of the general table of results.

4.- Presentation of two tables of results of Pythagoreans trios by parity (Even or Odd).

5.- Descriptive analysis. Behavior of Criterion and its application for the deduction of all and every

possible primitive Pythagorean trios.

6.- Conclusion: Presentation of the criterion


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Presentation of the Functional Equations or Formulas To Generate Every Possible

Pythagorean Trios.

Formula or functional equation according to parity:

For a = N = odd positive integer.

Then: N = a

When: (N2 - 1) / 2 = b

Where: ((N2 - 1) / 2) + 1 = c

For a = N = positive integer number.

Then: N = a

When: (N2 - 4) / 4 = b

Where: ((N2 - 4) / 4) + 2 = c


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Descriptive Analysis and Behavior of Criterion and Deduction of the Primitive

Pythagorean Trios.

72

= 49

1 48

2 47

3 46

4 45

5 44

6 43

7 42

8 41

9 40

10 39

11 38

12 37

13 36

14 35

15 34

16 33

17 32

18 31

19 30

20 29

21 28

22 27

23 26

24 25


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Remark: the prior scheme describes a decomposition of two addends of any square whose base is

an odd number.

72 = 49

1 48

2 47

3 46

4 45

5 44

. .

. .

25 12 37 49

. . .

9 20 29 49

. . .

1 24 25 49


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72 = 49

1 48

2 47

3 46

4 45

5 44

. .

. .

25 12 37 49 352 + 122 = 372

. . .

9 20 29 49 212

+ 202

= 292

. . .

1 24 25 49 72 + 242 = 252

. .

b c a2 + b2 = c2

Remark: During this development and analysis of criterion can be observed and inferred that the

terms of all primitive trios terms are obtained as follows: The terms (b and c) are specified in

columns by the decomposition of any square of any odd number into two addends and the term: (a)

is obtained with the product of the square roots of the differences and the sum of the terms of the

columns of the decomposition: (c -b). (b + c).


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Method to Obtain All Possible Pythagorean Trios:

Based: On the functional equations and formulas.

First step: Placing vertically all natural numbers. From: The number 1 to the number N.

1

2

3

4

5

6

.

.

.

N

Second step: We squared all of them.

12

22

32

42

52

62

.

.

.

N2


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Step Three: Place a second column vertically and parallel to the first by applying the appropriate

formula and according to the parity of each term in the first column and each term or result is also

squared.

12 + 02

22

+ 02

32 + 42

42

+ 32

52 + 122

62 + 82

. .

. .

. .

N2

b2

Remark: The terms of the second column are obtained according to the parity of the termcorresponding to the first column.

Step Four: Place vertically the terms of a third column applying the corresponding formulas

according to parity and squared after being separated by a plus sign and the sign of equality thus

obtaining all possible Pythagorean trios

12

+ 02

= 12

22 + 02 = 22

32

+ 42

= 52

42 + 32 = 52

52 + 122 = 132

62

+ 82

= 102

. .


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. .

. .

N2 + b2 = c2

a2 + b2 = c2 = Pythagorean equation

Descriptive Analysis and Deductions for the Characterization of Primitive Pythagorean Trios

12

+ 02

= 12

(Trivial)

2

2

+ 0

2

= 2

2

(Trivial)

32 + 42 = 52 (Primitive)

42

+ 32

= 52

(Primitive Permuted)

52

+ 122

= 132

(Primitive)

62 + 82 = 102 (Non-primitive)

. .

. .

. .

N2 + b2 = c2

a2

+ b2

= c2

= Pythagorean equation


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Presentation of the Criterion

Criterion: In order that a Pythagorean trio be a primitive Pythagorean trio is only necessary and

sufficient that: the second term is an even number when the third term is an odd number, where alsothe sum and difference of both terms are an exact square and the first term is the product of the

bases of both squares.

Criterion: Let: a2

+ b2

= c2

be a primitive Pythagorean trio.

It is only necessary and sufficient:

That (c - b) = x2

(an exact square)

When: (c + b) = y2 (an exact square)

Where: x. y = a (first term)

For: c = odd number (third term)

Where: b = number (second term)

General table of results

12

+ 02

= 12

(Trivial)

22 + 02 = 22 (Trivial)

32 + 42 = 52 (Primitive)

42

+ 32

= 52


52 + 122 = 132 (Primitive)

62 + 82 = 102 (Non-primitive)

. .

. .

. .

N2 + b2 = c2


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Two Tables of Results Depending on the Parity (Even and Odd)

Odd Examples:

12 + 02 = 12 (Trivial)

32

+ 42

= 52

(Primitive)

52

+ 122

= 132

(Primitive)

Even examples:

22 + 02 = 22 (Trivial)

42

+ 32

= 52


62

+ 82

= 102

(Non-primitive)

Remark: Separating the results by parity allows us to rule out non-primitive Pythagorean trios of

the primitive ones for the criterion conditions.


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Discussion and Recommendations:

Example of application:

theorem:

If n = Pi + PJ

When: Pi and Pj belong to the primes

Where: Pj - Pi = 2

Then: n / 2 = x

So that: (x2

+ 1)2

= a2

When: a2 - b2 = c2

Iff: b = n

When: (x2

- 1)2

= c2

Then: (x2 + 1) = a

When: (x2

- 1) = c


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Table of Results and Visualization of the Theorem:

Pi + Pj = n a2

- b2

= c2

3 + 5 = 8 172

- 82

= 152

5 + 7 = 12 372 - 122 = 352

11 + 13 = 24 1452 - 242 = 1432

Theorem: If the values for c are infinite. Then: The twin primes are infinite (see table for even

examples).

Remark: Apply the following general equation:

(((PI + PJ)/2)2 + ((PJPI)/2)

2)2 - (((PI + PJ).(PJPI))/2)2 = (PI . PJ)

2

Where: (((PI + PJ)/2)2

+ ((PJPI)/2)2))

2= a

2(First term)

When: (((PI + PJ).(PJPI))/2)2

= b2

(Second term)

For: (PI.PJ)2 = c2 (Third term)

Where: Pi Pj (belong to odd prime numbers)


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Theorem:

If: (a + b) belongs to the prime numbers.

When: (a - b) belongs to the prime numbers.

Then: (a2 - b2) is a semi-prime.

And if: (b / 2)2

+ 1 = a

Then: (a2b2) is product of two twin primes.

Where: (b / 2) + 1) belongs to the prime numbers.

When: (b / 2) - 1) belongs to the prime numbers.

Theorem: Every odd number can be expressed as the difference of two squares

12

- 02

= 1

22

-

12

= 3

32 - 22 = 5

42 - 32 = 7

32 - 02 = 52 - 42 = 9

62

- 52

= 11

72

- 62

= 13

42 - 12 = 82 - 72 = 15

92

- 82

= 17

102

- 92

= 19

52

- 22

= 112

- 102

= 21


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122

- 112

= 23

5202 = 132 - 122 = 25

62

- 32

= 142

- 132

= 27

152 - 142 = 29

162 - 152 = 31

72 - 42 = 172 - 162 = 33

62 - 12 = 182 - 172 = 35

192

- 182

= 37

82

- 52

= 202

- 192

= 39

212 - 202 = 41

222

- 212

= 43

72

- 22

= 92

- 62

= 232

- 222

= 45

242

- 232

= 47

72 - 02 = = 252 - 242 = 49

102 - 72 = 262 - 252 = 51

272 - 262 = 53

82 - 32 = 282 - 272 = 55

112 - 82 = 292 - 282 = 57

302 - 292 = 59

312

- 302

= 61

82

- 12

122

- 92

= 322

- 312

= 63

92

- 42

= 332

- 322

= 65

342

- 332

= 67

132

- 102

= 352

- 342

= 69


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Criterion:

All difference of squares of the form: (n + K)2- k

2

Is a semi-prime of the form: n2 + 2.n.K

If and only If: n belongs to prime numbers

When: (n + 2.K) belongs to prime numbers

Proof:

Since: (n + K)2 k2 = n2 + 2.n.K

For all: K0

Where: n1

When: n is greater than: K

And as: n2 + 2.n.K = n(n + 2.K)

Then: n(n + 2.K) is a semi-prime


When: (n + 2.K) belongs to prime numbers

And as: It is accomplished for all: n1

Where: K0

When: n is greater than: K

Then: All semi-prime is equal to the difference of two squares

Remark: Since all odd number can be expressed as the difference of two squares

And all semi-prime product of two odd prime numbers is an odd number


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Then all semi-prime can be expressed as the difference of two squares

And since the smallest odd prime number is number three

Then all semi-prime is the difference of two squares for all: n3

And since all semi-prime is the difference of two squares of the form:

(n + K)2- k

2

For all: n3

When: K0


When: (n + 2.K) also belongs to prime numbers

And since: semi-primes are infinite

Because: prime numbers are infinite

Then: For all: n belonging to prime numbers

There exist infinitely: (n + 2.K) also belonging to prime numbers

Where: n3

When: K0

Therefore: Polignacs conjecture is true

And since: All the above is accomplished for all: K0

And as also is accomplished when: K = 1

Then: Twin primes are infinite

And as: All semi-prime can be expressed as the difference of two squares

For all Para: n3

Then: Golbachs conjectures are both true


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Theorem: All odd prime number can be expressed of unique form as the difference of two

consecutive squares.

Corollary: All free of squares of (n) factors; can be expressed as the difference of two

squares of: 2n-1 different forms.

Corollary: All semi-prime can be expressed of two different forms as the difference of two

squares.

Theorem: There always exist: m < n

For all: n I

Such that: nm = I

Where: I is any odd number.

Theorem: There always exist: m < n

For all: n I

Such that: n + m = I

Where: I is any odd number

Theorem: I = n2m2

Theorem: I = Pi . Pj

Theorem: n2m2 = Pi . Pj

Theorem: (n + m).(n m) = Pi . Pj

Theorem: (n + m) + (nm) = 2.n = Pi .Pj

Theorem: If for all: n

There always exist: m

Such that: n2m2 = Pi . Pj

Where: Pi and Pj are both odd prime numbers

Then: Golbachs conjecture is true.


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Main Theorem:

n = n n = K n = n

If: 2.n - 1 = 2.n - 1 + 2.n - 1n = 1 n = 1 n = K + 1

K = Constant

For all: n 1

When: K 0

n = n n = K n = n

Then: 2.n - 1 - 2.n - 1 = 2.n - 1n = 1 n = 1 n = K + 1

K= Constant

Proof of Golbach Odd and Evens Conjectures by Reduction:

n=n

Since: n2

= 2.n1n=1

For all: n 1

n=K

And as: K2 = 2.n1n=1

For all: K 0


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n=n n=K

Then: n2 - K2 = 2.n1 - 2.n1n=1 n=1

n=n n=K n=n

Where: = 2.n1 - 2.n1 = 2.n - 1n=1 n=1 n=K + 1

For all: n1

When: K 0

n=n

Then: n2 - K2 = 2.n - 1n = K + 1

And as: n2

- K2

= (nK). (n + K) (By theorem 1)

n=n

Then: (nK) (n + K) = 2.n - 1 = Semi-primen = K + 1

If and only If: (n + K) is the central term of:

n = n

2.n - 1 = Semi-primen = K + 1

When: (n - K) is the number of terms of:

n = n

2.n - 1 = Semi-primen = K + 1


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Where: (nK) belongs to prime numbers.

When: (n + K) belongs to prime numbers.

Theorem:

If: n = Ii + K

When: n = Ij - K

Then: 2.n = Ii + Ij

And as: (nK)(n + K) = Ii . Ij = n2 - K2

For all: n 3

When: K0

Where: n < K

Then: Ij (n K) n (n + K) Ii

For all: Odd number: Ij

And all: Odd number: Ii


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Theorem: For all: Odd number of the form: Ii.Ij

There always exist: Two numbers of the form: (nK) y (n + K)

Such that: (nK) = Ii

When: (n + K) = Ij

For all: n > K

Where: Ii is any odd number

When: IJ is any odd number.

Then: (nK)(n + K) = n2K2 = Ii . Ij

Theorem: All odd multiples smallest of any odd number squared can be expressed as the

difference of two squares.

22 - 12 = 3

32

- 02

= 9

32

- 22

= 5

42

- 12

= 15

52 - 02 = 25

42 - 32 = 7

52

- 22= 21

62 - 12 = 35

72

- 02= 49


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Main Theorem of Existence:

For all: n 3

There always exist: (n) numbers less than or equal to: n2

Between: 2.n1 and n2

Which are: Product of two equidistant numbers to: n

Of the form: (nK) and (n + K)

Where: 0 K n

Such that: (nK)(n + K) = n2 - K2

Where: at least one of these: (n) Products is: A semi-prime.

Where: The sum of both factors is equal to: (nK) + (n + K) = 2.n

Example:

n2

- K2

= ( nK ).( n + K ) = Products 2.n = ( nK ) + ( n + K ) = Sums

72 - 02 = 7 . 7 = 49 7 + 7 = 14

72 - 12 = 6 . 8 = 48 6 + 8 = 14

72 - 22 = 5 . 9 = 45 5 + 9 = 14

72 - 32 = 4 . 10 = 40 4 + 10 = 14

72

- 42

= 3 . 11 = 33 3 + 11 = 14

72

- 52

= 2 . 12 = 24 2 + 12 = 14

72

- 62

= 1 . 13 = 13 1 + 13 = 14


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n2

- K2

= (nK).(n + K) = Products

72

- 02

= 7 . 7 = 49..= n2

72

- 12

= 6 . 8 = 48

72 - 22 = 5 . 9 = 45

72 - 32 = 4 . 10 = 40

72 - 42 = 3 . 11 = 33. = Semi-prime

72 - 52 = 2 . 12 = 24

72

- 62 = 1 . 13 = 13. = 2.n 1

1

3 4

5 8 9

7 12 15 16

9 16 21 24 25

11 20 27 32 35 36

13 24 33 40 45 48 49Example

15 28 39 48 55 60 63 64

17 32 45 56 65 72 77 80 81


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Theorem:

All nth. prime number: Pn can be expressed as: (n) pair of numbers less or equal than: Pn

such that: the sum and the difference of both be a prime number.

Example: Difference (nK) = 1 3

K 1 0

Relationship:

n 2 3

Sum (n + K) = 3 3Pn

Example: Difference.. (n K) = 1 3 5

K 2 1 0

Relationship:

n 3 4 5

Sum (n + K) = 5 5 5Pn

Example: Difference (nK) = 1 3 5 7

K 3 2 1 0

Relationship:

n 4 5 6 7

Sum: (n + K) = 7 7 7 7Pn

Example: Difference: (nK) = 1 3 5 7 9 11 13 15 17 19

K 9 8 7 6 5 4 3 2 1 0


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Relationship:

n 10 11 12 13 14 15 16 17 18 19

Sum: (n + K) = 19 19 19 19 19 19 19 19 19 19Pn

Justification: For all nth odd prime number: Pn There always exist (n) odd prime numbers

less or equal than: Pn

Example: Difference: (nK) = 1 3 5 7 9 11 13 15 17 19K 9 8 7 6 5 4 3 2 1 0

Relationship:

n 10 11 12 13 14 15 16 17 18 19

Sum:(n + K) = 19 19 19 19 19 19 19 19 19 19P7

Application: On Golbachs even conjeture: Since for all nth odd prime number: Pn therealways exist: (n) even numbers that can be expressed as the sum of two prime numbers and

since the primes are infinite. Then: there always exist infinitely even numbers where the

conjecture is true.

Remark: 2.n is sum of two odd prime numbers of the form: (nK) + (n + K)


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Presentation of Criteria of Primality

Criterion 1: For all: n

There always exist infinitely arithmetic progressions of reason: 2.n + 1

Where: The first term is: 2.n2

+ 2.n + 1

Which: do not express infinitely positive numbers of the form: K

Such that: 2.K1 is an odd prime number.

Criterion 2: For all: n

There always exist infinitely arithmetic progressions of reason: 2.n + 1

Where: The first term is: 2.n2

+ 2.n

Which: do not express infinitely positive numbers of the form: K

Such that: 2.K + 1 is an odd prime number.


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References

[1] Rada deMatematicas.Elementales.(Aritmetica).Cenamec.1992.Caracas.pag.42-44

[2]Joice.D 2005 (Consulta en Internet el 3/11/2011)

[3]Enciclopedia tematica Espasa.1998.Espaa.pag.459-460

[4]Nieves Rivas.R. Prueba de Primalidad.XVIII Jornadas de Investigacion y II de

Postgrado.Memorias de la Unellez.Venezuela.2011.pag.216-

[5]Nieves Rivas R. Demostracion de una conjetura presentada en el quinto congreso de

Matematicas en 1912. XIX Jornadas de Investigacion y III de Postgrado.Memorias de la

Unellez.Venezuela.2011

Documents

Methodological Strategy for Obtaining the Possible Pythagoreans Trios and Proof Through a Criterion Applied to the Primitive Pythagorean Trios