User:Springelec/Pythagorean Triples

'New Pythagorean Triple Formula'

1.0) Introduction.

In 1985, the BBC’s flag ship science programme, Horizon, broadcast a programme about unsolved theories and problems which included Fermat’s Last Theorem of 1637 in which Fermat* stated that no integer solutions existed for the equation:

aⁿ = bⁿ + cⁿ, (Eq 1); if n was an integer whose value was greater than 2.

Intrigued by the simplicity of the equation and the fact that, at the time, no one had found or rediscovered Fermat’s alleged general proof, the author made various attempts to solve Fermat’s Last Theorem. Although no claim is made to have either rediscovered Fermat’s proof nor offer an alternative, analysis of Fermat’s original problem has allowed a novel technique to be formulated which allows the automatic generation of Pythagorean triples from a seed integer. Examples include 5² = 4² + 3² and 13² = 12² + 5².

2.0) History.

It should be noted that this abstract equation is a multi dimensional equation depending upon the value of ‘n’ and refers to a right angled triangle when n = 2, as illustrated below for clarity. In this example sides A, B & C are mathematically related by the equation A² = B² + C², i.e. aⁿ = bⁿ + cⁿ when n = 2.

Fig 1

The reader will quickly appreciate that when n = 1, (aⁿ = bⁿ + cⁿ ) is reduced to just (a = b + c), an expression which has an infinite number of real number and integer solutions. The reader will remember from school maths lessons that when n = 2, (aⁿ = bⁿ + cⁿ ) becomes the famous Pythagorean equation for right angled triangles, (a² = b² + c² ) from which can be calculated the hypotenuse of a right angled triangle providing two lengths or one length and an angle are known.

Famous Pythagorean triples include 5² = 4² + 3² and 13² = 12² + 5², triples which were known to the ancient Sumerians and Babylonians four thousand years ago because they used the formula (p² – q²), 2pq, (p² + q²) to calculate perfect triples. The reader can quickly determine this formula works by substituting p = 2, q = 1 or p = 3, q = 2 for example, but it is all the more remarkable that the Babylonians also worked out that 2291, 2700 and 3541 is also a perfect triplet using this method.

It is not known why the Sumerians and Babylonians needed to know these perfect triples, other than for, perhaps, agricultural mapping, but regardless of the exact reason, several perfect integer triples have been known to man for at least four thousand years and knowledge of their existence has always fascinated mankind for practical and aesthetic reasons.

Pierre de Fermat is the first person recorded in history to seek a general solution for the abstract equation aⁿ = bⁿ + cⁿ and whilst solutions for n = 1 and n = 2 were known for a long time, it would have seemed odd to Pierre Fermat and other mathematicians throughout history that there appeared to be no known integer solutions for the cubic volume equation, a³ = b³ + c³ or the fourth order space equation, a⁴ = b⁴ + c⁴ or even higher order triples. Pierre Fermat was curious to understand this problem and went on to prove that there were no integer solutions for the two specific equations when n = 3 and n = 4. Whilst the proof for n = 3 is lost, n = 4 is extant and provided an insight into Fermat’s analytical methods. His proof was based upon the assumption that three integer solutions existed to satisfy the equation a⁴ = b⁴ + c⁴. The analysis went on to show that three smaller integers could also exist which also satisfied the equation and so on, down to the smallest set of a, b and c which clearly did not satisfy the equation. Once a minimum set of a, b and c were found that could not satisfy the equation, then none of the previous values for a, b and c could satisfy the equation because all sets of a, b and c were mathematically linked. This mathematical strategy aimed to prove a flaw or logical inconsistency with the original argument and by proving the flaw, bring down the whole argument like a house of cards. In this way, Fermat proved there were no integer solutions for a⁴ = b⁴ + c⁴.

Although lost, there is little doubt Fermat would have used the same argument to also prove there are no integer solutions for when n = 3. One of the reasons why Fermat’s proof for n = 3 is lost is because Fermat didn’t keep very good records of his mathematical research, preferring to scribble notes in the margins of other people’s work which he read for enjoyment. Whilst working on a general solution for the equation (aⁿ = bⁿ + cⁿ ) and translating and editing one of the five extant copies of thirteen books of Arithmetica, written by the Greek mathematician, Diophantus, Fermat is said to have written in a margin: “I have found a truly marvellous demonstration which this margin is too narrow to contain”. The demonstration Fermat referred to was a proof that there were no integer solutions for the abstract equation, (aⁿ = bⁿ + cⁿ ) for all integer values of n > 2. Unfortunately, Fermat either forgot to record his thoughts on this topic or they were lost after his death. Either way, the World was only left with his tantalising note suggesting that he had solved this problem with a seemingly simple and elegant proof.

Whether Fermat did or did not prove that there are no integer solutions for (aⁿ = bⁿ + cⁿ ) for all values of n > 2 remains a mystery due to the loss of much of his work. History granted Fermat the benefit of doubt for this discovery because he had already proven many theorems and solved the problem for the two discrete cases when n = 3 and n = 4. However, the tantalising note Fermat left behind in 1637, frustrated the attempts of very many well known mathematicians to rediscover Fermat’s alleged proof, all without success. French mathematician Legendre proved there were no integer solution for n = 5 in 1823 and in 1840, Lamé and Lebesgue gave proofs for when n = 7. Dirichlet gave a proof for n = 14 and in 1849, Kummer proved that there were no integer solutions for some Bernoullian conditions, 37, 59 and 67 being exceptions. Because Fermat’s last theorem defied solution for nearly 360 years, it acquired an extraordinary celebrity status among mathematicians until Princeton mathematician, Professor Andrew Wiles indirectly, but finally proved the theorem in 1995 as a consequence of proving the Taniyama Shimura conjecture.

3.0) A new analytical approach. Although Andrew Wiles proved Fermat’s Last Theorem, the following analysis is offered as a bi-product of the authors unsuccessful attempts to solve Fermat’s Last Theorem.

Given the right angled triangle below, we know from Pythagoras that the sides A, B and C are related by the equation; (A² = B² + C² ) and through the centuries many perfect Pythagorean triples have been found, (5² = 4² + 3² ) and (13² = 12² + 5² ) are just two of the most recognisable triples.

Table 1 illustrates a sample of eleven integer triples, although there are infinitely many more.

Fig 2

A B C 5 4 3 13 12 5 17 15 8 25 24 7 29 21 20 37 35 12 41 40 9 61 60 11 65 63 16 85 84 13 113 112 15

Table 1

Ancient mathematicians looked for patterns and symmetry in numbers to find an underlying truth. Using this philosophical approach, we begin this analysis of Pythagorean triples by assuming there is an underlying relationship between the numbers that make up all Pythagorean triples. Since the hypotenuse of all right angles triangles is always the largest number, we start by converting the Pythagorean equation:

(A² = B² + C²) into an equivalent that links the three numbers together, i.e. A² = (A-p)² + (A-q)²

For clarity, we also choose to replace A with n, such that the equation becomes:

n² = (n – p)² + (n – q)² ………….(Eqn 2)

Expanding terms, we get: n² = n² – 2np + p² + n² – 2nq + q²

Collecting terms, we get: n² = 2n² – 2n(p + q) + (p² + q²)

And hence from this analysis we form a new quadratic equation: n² – 2n(p + q) + (p² + q²) = 0

Using the standard quadratic formula to solve this equation: (-b ± √(b² – 4ac))/2a

The solutions are n₁, n₂ = (p + q) ± √2pq ………….(Eq 3)

The difficulty with solving this solution for p and q is the surd, ± √2pq which restricts the number of possible integer solutions. To find this limit, we need to ensure ± √2pq becomes a perfect square, i.e. the values of p and q always ensure 2pq is a perfect square and therefore always produces an integer square root, e.g. 4 roots to 2, 9 roots to 3, 16 roots to 4 and so on.

Let r = ± √2pq, then r² = 2pq and making p the subject: p = r² / 2q

From which it is seen that for all values of q, the denominator, 2q is always even.

This means ‘r’ cannot be an odd number because r² / 2q will always leave a fractional remainder.

This can be shown as follows:

Let ‘r’ be any generalised odd number, i.e., r = (2u + 1) where ‘u’ is any integer.

Therefore: r² = (2u + 1)(2u + 1) and hence p = (2u + 1)(2u +1)/ 2q, i.e. p = (4u2 + 4u + 1)/2q

Any value of ‘u’ will always leave a remainder equal to 1/2q, which is fractional because q is defined to be an integer. Therefore for a perfect square to exist, ‘r’ cannot be an odd number. Since all numbers must be either odd or even, ‘r’ must therefore be an even integer, i.e.

r = 2u ► p = (2u)² / 2q ► p = 4u² / 2q ► 2pq = 4u² ► pq = 2u²

and we have achieved the goal of expressing 2pq in terms of a perfect square, since 4u² will generate perfect integer roots for all values of ‘u’. Table 2 is a small tabulation of the first ten positive integers to illustrate the effect.

u 4u² ± √2pq = ±√4u² 0 0 0 1 4 2 2 16 4 3 36 6 4 64 8 5 100 10 6 144 12 7 196 14 8 256 16 9 324 18 10 400 20

Table 2

Note that ‘u’ represents any positive, zero or negative seed integer and therefore p and q represent integer factors which can be used to solve the quadratic solution; n1, n2 = (p + q) ± √2pq and because steps have been taken to ensure 2pq is a perfect square, n₁ and n₂ will therefore always exist as integers.

Factors p and q are actually derived from 2u² because there is a redundant 2 in the 2pq = 4u² relationship. Factors p and q are therefore calculated from pq = 2u²

4.0) Application. Because the seed integer ‘u’ is squared, ’u’ can be any positive, zero or a negative integer, but for simplicity let u = 1 to begin.

Let u = 1 2u² = 2(1)² = 2. Therefore, non repeating factors of 2u² are p = 1 q = 2

Note that once the pq factors repeat, the process of factorising the integer, 2u² stops because this would produce redundant factors and therefore redundant information. Using the quadratic solution from page 5, the two possible values of n are:

n₁, n₂ = (p + q) ± √2pq ………..(Eq 3)

i.e. n₁, n₂ = (1 + 2) ± √2(1)2

n₁, n₂ = 3 ± 2 i.e. n₁, n₂ = 5 or 1

Substituting these integers into Eqn 2, we get:

n₁² = (n₁ - p)² + (n₁ - q)² or n₂² = (n₂ - p)² + (n₂ - q)²

5² = (5 – 1)² + (5 – 2)² or 1² = (1 – 1)² + (1 – 2)²

5² = 4² + 3² or 1² = 0² + (-1)²

And we discover that when the seed integer is unity, two Pythagorean triples are created, but only one is of any real interest. The other is known as a trivial solution because it equates a number to itself and contains zero in the solution.

Let u = 2 2u² = 2(2)² = 8 and the non repeating factors of 2u² are: p = 1 2 q = 8 4

Now we have four factors when u = 2 which will generate four Pythagorean triples.

Substituting these factors into Eqn 3, we get:

n₁, n₂ = (1 + 8) ± √2(1)8 and n₃, n₄ = (2 + 4) ± √2(2)4

n₁, n₂ = 9 ± 4 and n₃, n₄ = 6 ± 4

n₁, n₂ = 13 or 5 and n₃, n₄ = 10 or 2

Substituting these integers into Eqn 2, we get:

n₁² = (n₁ – p₁)² + (n₁ – q₁)² & n₂² = (n₂ – p₁)² + (n₂ – q₁)²

& n₃² = (n₃ – p₂)² + (n₃ – q₂)² & n₄² = (n₄ – p₂)² + (n₄ – q₂)²

13² = (13 – 1)² + (13 – 8)² & 5² = (5 - 1)² + (5 - 8)²

& 10² = (10 – 2)² + (10 – 4)² & 2² = (2 - 2)² + (2 - 4)²

13² = 12² + 5² & 5² = 4² + (-3)² & 10² = 8² + 6² & 2² = 0² + (-2)²

And we discover that when the seed integer, u = 2, it generates three real and one trivial Pythagorean triples, although strictly speaking only one new triple has been calculated because the (5² = 4² + 3²) triple has been repeated twice in the form of (5² = 4² + (-3)²) and (10² = 8² + 6²).

Let u = 3 2u² = 2(3)² = 18 and the non repeating factors of 2u² are: p = 1 2 3 q = 18 9 6

Now we have six separate possible solutions when u = 3, i.e. 2 x the non repeating factors of pq.

Substituting these integers into Eqn 3, we get:

n₁, n₂ = (1 + 18) ± √2(1)18 & n₃, n₄ = (2 + 9) ± √2(2)9 & n₅, n₆ = (3 + 6) ± √2(3)6

n₁, n₂ = 19 ± 6 & n₃, n₄ = 11 ± 6 & n₅, n₆ = 9 ± 6

n₁, n₂ = 25 or 13 & n₃, n₄ = 17 or 5 & n₅, n₆ = 15 or 3

Substituting these integers into Eq 2, we get:

n₁, n₂ = n² = (n–p₁)² + (n–q₁)² & n₃, n₄ = n² = (n–p₂)² + (n–q₂)² & n₅, n₆ = n² = (n–p₃)² + (n–q₃)²

25² = (25 – 1)² + (25 – 18)² & 17² = (17 - 2)² + (17 - 9)² & 15² = (15 – 3)² + (15 – 6)²

25² = 24² + 7² & 17² = 15² + 8² & 15² = 12² + 9²

and

13² = (13 – 1)² + (13 – 18)² & 5² = (5 - 2)² + (5 - 9)² & 3² = (3 – 3)² + (3 – 6)²

13² = 12² + (-5)² & 5² = 3² + (-4)² & 3² = 0² + (-3)²

Of these six results, only two are new triples, (25² = 24² + 7² ) and (17² = 15² + 8² ). The 13² = 12² + (-5)² triple is a version of the 13² = 12² + 5² triple found when u =2. The (15² = 12² + 9² ) triple is just three times the 5² = 4² + 3² triple found when u =1. The 5² = 3² + (-4)² triple is a variant of the 5² = 4² + 3² triple calculated when u = 1. Finally, the last triple is the trivial solution 3² = 0² + (-3)² which just equates a number to itself with the addition of zero.

5.0) Pythagorean Patterns of Behaviour. There now follows a list of the Pythagorean triples calculated from seed values of ‘u’ between 1 and 10 to illustrate the patterns that exist in these integers. A Basic programme was written for this task. The two most interesting patterns that emerged quite quickly from this limited study were:

1) Each triple was made from two odd integers and one even integer. 2) The area of the triangle and the sum of the triples were related to the seed integer, ‘u’. 3) The seed integer has the dimension of ‘metres’.

Both these topics are discussed later in this paper, but expressed mathematically, it will be shown that:

Seed integer, u = (2 x Triangle Area / Sum of triples) ► u = (BC / (A+B+C))

Using this new fact, it has been possible to calculate the seed integer for the impressive Sumerian and Babylonian triple, A = 3541, B = 2700, C = 2291 was 725. Naturally, they would have been completely unaware of the existence of this seed integer, nor the other triples that this number produced. A full catalogue of the ‘725’ triples is reproduced at the end of this section.

In the following tabulation, each seed integer generates a number of Pythagorean triples, some unique, some variants, but always one trivial triple. For the purposes of clarity, each triple is described by these terms.

u = 1

5² = 4² + 3² Unique 1² = 0² + (-1)² Trivial

There are two solutions, including the trivial solution is; u² = 0² + (-u)²

u = 2

13² = 12² + 5² Unique 5² = 4² + (-3)² Variant 10² = 8² + 6² (5u)² = (4u)² + (3u)² ) 2² = 0² + (-2)² Trivial

Although there are four solutions, only one is original. The 5² = 4² + (-3)² triple is a variation of 5² = 4² + 3² & 10² = 8² + 6² is (5u)² = (4u)² + (3u)². The trivial solution is again u² = 0² + (-u)²

u = 3

25² = 24² + 7² Unique 17² = 15² + 8² Unique 15² = 12² + 9² (5u)² = (4u)² + (3u)² 13² = 12² + (-5)² Variant 5² = (-4)² + 3² Variant 3² = 0² + (-3)² Trivial

There are six (2u) solutions, but only two are original. 13² = 12² + (-5)² and 5² = 3² + (-4)² are variants of 13² = 12² + 5² and 5² = 4² + 3² and 15² = 12² + 9² is just (5u)² = (4u)² + (3u)² . Note again the trivial solution is also u² = 0² + (-u)²

u = 4

41² = 40² + 9² Unique 25² = 24² + (-7)² Variant 26² = 24² + 10² 2 x (13² = 12² + 5²) 10² = 8² + (-6)² 2 x (5² = 4² + (-3)²) 20² = 16² + 12² (5u)² = (4u)² + (3u)² 4² = 0² + (-4)² Trivial

u = 5

61² = 60² + 11² Unique 37² = 35² + 12² Unique 41² = 40² + (-9)² Variant 17² = 15² + (-8)² Variant 25² = 20² + 15² (5u)² = (4u)² + (3u)² 5² = 0² + (-5)² Trivial

u = 6

85² = 84² + 13² Unique 29² = 21² + 20² Unique 61² = 60² + (-11)² Variant 5² = (-4)² + (-3)² Variant 39² = 36² + 15² 3 x (13² = 12² + 5²) 15² = 12² + (-9)² 3 x (5² = 4² + (-3)²) 50² = 48² + 14² 2 x (25² = 24² + 7²) 26² = 24² + (-10)² 2 x (13² = 12² + (-5)²) 34² = 30² + 16² 2 x (17² = 15² + 8²) 10² = (-8)² + 6² 2 x (5² = 4² + 3²) 30² = 24² + 18² (5u)² = (4u)² + (3u)² 6² = 0² + (-6)² Trivial

u = 7

113² = 112² + 15² Unique 65² = 63² + 16² Unique 85² = 84² + (-13)² Variant 37² = 35² + (-12)² Variant 35² = 28² + 21² (5u)² = (4u)² + (3u)² 7² = 0² + (-7)² Trivial

u = 8

145² = 144² + 17² Unique 113² = 112² + (-15)² Variant 82² = 80² + 18² 2 x (40² = 40² + 9²) 50² = 48² + (-14)² 2 x (25² = 24² + (-7)²) 52² = 48² + 20² 4 x (13² = 12² + 5²) 20² = 16² + (-12)² 4 x (5² = 4² + (-3)²) 40² = 32² + 24² (5u)² = (4u)² + (3u)² 8² = 0² + (-8)² Trivial

u = 9

181² = 180² + 19² Unique 145² = 144² + (-17)² Variant 101² = 99² + 20² Unique 65² = 63² + (-16)² Variant 75² = 72² + 21² 3 x (25² = 24² + 7²) 39² = 36² + (-15)² 3 x (13² = 12² + (-5)²) 51² = 45² + 24² 3 x (17² = 15² + 8²) 15² = (-12)² + 9² 3 x (5² = 4² + 3²) 45² = 36² + 27² (5u)² = (4u)² + (3u)² 9² = 0² + (-9)² Trivial

u = 10

221² = 220² + 21² Unique 181² = 180² + (-19)² Variant 122² = 120² + 22² 2 x (61² = 60² + 11²) 82² = 80² + (-18)² 2 x (41² = 40² + (-9)²) 74² = 70² + 24² 2 x (37² = 35² + 12²) 34² = 30² + (-16)² 2 x (17² = 15² + (-8)²) 65² = 60² + 25² 5 x (13² = 12² + 5²) 25² = 20² + (-15)² 5 x (5² = 4² + (-3)²) 53² = 45² + 28² Unique 13² = (-12)² + 5² Variant 50² = 40² + 30² (5u)² = (4u)² + (3u)² 10² = 0² + (-10)² Trivial

The Sumerian & Babylonian Seed Integer: 725 1052701² = 1052700² + 1451² Unique 527077² = 527075² + 1452² Unique 42341² = 42340² + 291² Unique 21317² = 21315² + 292² Unique 3757² = 3132² + 2075² Unique 3625² = 2900² + 2175² (5u)² = (4u)² + (3u)² 3541² = 2700² + 2291² Sumerian & Babylonian 1997² = 1972² + 315² Unique 1741² = 1740² + 59² Unique 1301² = 1300² + 51² Unique 1181² = 1131² + 340² Unique 901² = 899² + 60² Unique 725² = 0² + (-725)² Trivial 677² = 675² + 52² Unique 61² = 60² + 11² Variant 37² = 35² + 12² Variant 1049801² = 1049800² + (-1449)² Variant 524177² = 524175² + (-144811)² Variant 41761² = 41760² + (-289)² Variant 20737² = 20735² + (-288)² Variant 1625² = 1624² + (-57)² Variant 1417² = 1392² + (-265)² Variant 1201² = 1200² + (-49)² Variant 857² = (-825)² + 232² Variant 785² = 783² + (-56)² Variant 641² = (-609)² + (-200)² Variant 601² = 551² + (-240)² Variant 577² = 575² + (-48)² Variant 41² = 40² + (-9)² Variant 17² = 15² + (-8)² Variant

References

Author: Greg Spring. C. Eng B.Sc MIEE. (Institute of Electrical Engineers) Revision 1: Date: 20th January 2011. Revision 2: Date: 7th February 2011. Sections 2, 5, 6 & 7 amended. Revision 3: Date: 12th February 2011. Dimension statement for seed integer added.

• [Pierre de Fermat. Born Montauban, France, 1601. Died Castres, France, 12th January 1665.]

The reference for historical information was obtained from “A short account of the history of mathematics” by W. W. Rouse Ball, Trinity College, Cambridge, an unabridged internet copy of the author’s fourth edition published in 1908. International Standard Book Number: 0-486-20630-0. Library of Congress Catalog Card Number: 60-3187

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