Pythagorean Triples

Alexander Bisant, Howard Community College 2016, University of Maryland Baltimore County
Kyle Sutton, Howard Community College 2016, University of Maryland Baltimore County

Mentored by: Mike Long, Ed.D. & Loretta FitzGerald Tokoly, Ph.D.

Abstract

It was noticed while observing a specific subset of the Pythagorean Triples, that Triangular Numbers, positive integers which are the sum of the first “n” positive integers, were implicitly appearing in the Pythagorean Triples.  This led to the question and proof that Triangular Numbers can be used to generate a subset of the Primitive Pythagorean Triples.  An immediate extension was if Triangular Numbers could be used to generate all of the Primitive Pythagorean Triples.  A method was found for generating all of the Primitive Pythagorean Triples, along with some non-Primitive Triples as well, using an argument that actually ties back to the work of Euclid already mentioned.  A second argument can be made to filter out those Triples that are not primitive.

 

Introduction

Pythag1.PNGSince the time of the Plimpton 322 (1800 B. C.) when they were discovered etched in clay tablets, Pythagorean Triples, sets of three positive integers AB, and C, where LaTeX: A^2+B^2=C^2, have captured the interest of mathematicians.  Migrating many years later and many miles away, Euclid came up with an elegant proof to generate all of the Primitive Pythagorean Triples.  These are Triples where the mentioned integers, a, b, and c, have a greatest common divisor of one. That interest in Pythagorean Triples has migrated around the globe and has been mentioned in the work of countless mathematicians.

Even today, interest in and exploration of Pythagorean Triples, which is the focus of this research, remains alive. Initially, it was observed with Pythagorean Triples in the list shown (see Figure 1), that each of the middle numbers, which represents the length of the longer leg of the right triangle, is always four times a Triangular Number. It was also observed that the shorter leg was an odd number and the hypotenuse was one more than the longer leg. The research question that emerged was whether all of the Primitive Pythagorean Triples could be generated using the Triangular Numbers in one way or another.

Background

A Pythagorean Triple is a set of three unique positive integers LaTeX: AALaTeX: BB, and LaTeX: CC for which the following is true:

LaTeX: A^2+B^2=C^2

where LaTeX: AA and LaTeX: BB are the legs of a right triangle and LaTeX: CC is the hypotenuse of that right triangle. For a triple to be considered “primitive”, the three integers must be pairwise relatively prime, or each pair has a greatest common divisor of LaTeX: 11LaTeX: gcd(A,B)=gcd(A,C)=gcd(B,C)=1.

Probably the most well-known method for generating a complete covering of primitive Pythagorean Triples is the formula first introduced by Euclid around 300 BC. Euclid’s formula involves two positive, relatively prime integers, LaTeX: mm and LaTeX: nn, where imagen” src=”https://howardcc.instructure.com/equation_images/m%253En” alt=”LaTeX: m>n” data-equation-content=”m>n” data-mathml=” m > n “>m > n and LaTeX: m-nm − n is odd (in other words, one of the integers is odd and the other even).

Euclid’s formula defines the three sides of the triangle, LaTeX: ALaTeX: B, and LaTeX: C , as:

LaTeX: A=m^2-n^2~~~~~~~~~~~~~~~B=2mn~~~~~~~~~~~~~~~C=m^2+n^2

Initial Research Findings

For the infinite set of primitive triples, there exists a subset of primitive triples for which the difference of LaTeX: CC and LaTeX: BB is one. This subset of primitive triples is listed in ascending order based on the value of LaTeX: AA in Figure 2. The list of triples for LaTeX: n=1n = 1 through LaTeX: n=5n = 5 is identical to the list of triples in Figure 1 that led to the initial question.

LaTeX: mm LaTeX: nn LaTeX: A=m^2-n^2 LaTeX: B=2mn LaTeX: C=m^2+n^2
2 1 3 4 5
3 2 5 12 13
4 3 7 24 25
5 4 9 40 41
6 5 11 60 61
LaTeX: m LaTeX: m-1 LaTeX: 2m-1 LaTeX: 2m^2-2m LaTeX: 2m^2-2m+1
LaTeX: n+1 LaTeX: n LaTeX: 2n+1 LaTeX: 4\left(\frac{n(n+1)}{2}\right) LaTeX: 4\left(\frac{n(n+1)}{2}\right)+1

It should be noted that all of the ’s in Figure 2 are all multiples of 4. It can be easily shown that LaTeX: B=2m^2-2m=2m(m-1) is always a multiple of four, where LaTeX: mm is a natural number greater than or equal to 2. Since either LaTeX: mm or LaTeX: m-1m − 1 must be even, either LaTeX: 2m or LaTeX: 2(m-1) is a product of 2 and an even number. Therefore, LaTeX: BB is a multiple of four because the product of 2 and any even number is a multiple of four. Consequently, each LaTeX: BB in the ordered subset of primitive triples represented by Figure 2 can be divided by four and results in the following sequence: 1,3,6,10,15,… This sequence is the same as the sequence produced by the formula LaTeX: \frac{n(n+1)}{2}, which generates the sum of the first  natural numbers or the Triangular Numbers.


Further Research Findings

Pythag2.PNGThe first subset examined was defined by the difference of LaTeX: CC and  LaTeX: AA being 2 and is in Figure 3.

The triples of this set have the following properties for LaTeX: k\in Wk ∈ W:

LaTeX: B_k=4(1)+4(k); ; 4 times a triangular number, LaTeX: 1=\frac{(\color{red}1\color{black})((\color{red}1\color{black})+1)}{2}, plus a multiple of LaTeX: 4=4(\color{red}{1}\color{black}))LaTeX: A_k=(2k+2)^2-1=(2k+1)((2k+1)+2); the product of two odd numbers that differ by LaTeX: 2=2(\color{red}1\color{black});2 = 2 ( 1 ) ; LaTeX: C_k=(2k+2)^2+1=A_k+2(1); LaTeX: A_k plus 2 times a square number, LaTeX: 1=\color{red}1\color{black}^2.

 

 

Pythag3.PNG

The second subset examined was defined by the difference of LaTeX: CC and LaTeX: AA being 8 and is in Figure 4:

The triples of this set have the following properties for LaTeX: k\in W:k ∈ W :

LaTeX: B_k=4(3)+8(k); 4 times a triangular number, LaTeX: 3=\frac{(\color{red}2\color{black})((\color{red}2\color{black})+1)}{2} plus a multiple of LaTeX: 8=4(\color{red}2\color{black});LaTeX: A_k=(2k+3)^2-4=(2k+1)((2k+1)+4); the product of two odd numbers that differ by LaTeX: 4=2(\color{red}2\color{black});LaTeX: C_k=(2k+3)^2+4=A_k+2(4);LaTeX: A_kA k plus 2 times a square number, LaTeX: 4=\color{red}2\color{black}^2

 

 

 

The third subset examined was defined by the difference of LaTeX: C and LaTeX: A being 18 and is in Figure 5.

Pythag5.PNG

 

The triples of this set have the following properties for LaTeX: k\in W::

LaTeX: B_k=4(6)+12(k); 4 times a triangular number, LaTeX: 6=\frac{(\color{red}3\color{black})((\color{red}3\color{black})+1)}{2}, plus a multiple of LaTeX: 12=4(\color{red}3\color{black})LaTeX: A_k=(2k+4)^2-9=(2k+1)((2k+1)+6);A k = ( 2 the product of two odd numbers that differ by LaTeX: 6=2(\color{red}3\color{black});6 = 2 ( 3 ) ; LaTeX: C_k=(2k+4)^2+9=A_k+2(9); LaTeX: A_kA k plus 2 times a square number, LaTeX: 9=\color{red}3\color{black}^29 = 3 2.

 

 

 

It should be noted that this method does pick up some Pythagorean Triples that are not primitive.  In Figure 5, the triple 27, 36, 45 is not primitive.  But, is a multiple (9 times) of the Primitive Pythagorean Triple 3, 4, 5.

 

Pythag67.PNGThe fourth subset examined was defined by the difference of LaTeX: C and LaTeX: A being 32 and is in Figure 6.

The triples of this set have the following properties for LaTeX: k\in W:k ∈ W :

LaTeX: B_k=4(10)+16(k); 4 times a triangular number, LaTeX: 10=\frac{(\color{red}4\color{black})((\color{red}4\color{black})+1)}{2}, plus a multiple of LaTeX: 16=4(\color{red}4\color{black});  LaTeX: A_k=(2k+5)^2-16=(2k+1)((2k+1)+8); the product of two odd numbers that differ by LaTeX: 8=2(\color{red}4\color{black}); LaTeX: C_k=(2k+5)^2+16=A_k+2(16); LaTeX: A_kA k plus 2 times a square number, LaTeX: 16=\color{red}4\color{black}^2.

The fifth subset examined was defined by the difference of LaTeX: CC and LaTeX: AA being 50 and is in Figure 7.

The triples in this set have the following properties for LaTeX: k\in W:

LaTeX: B_k=4(15)+20(k); 4 times a triangular number, LaTeX: 15=\frac{(\color{red}5\color{black})((\color{red}5\color{black})+1)}{2}, plus a multiple of LaTeX: 20=4(\color{red}5\color{black});LaTeX: A_k=(2k+6)^2-25=(2k+1)((2k+1)+10);the product of two odd numbers that differ by LaTeX: 10=2(\color{red}5\color{black});1 LaTeX: C_k=(2k+6)^2+25=A_k+2(25); LaTeX: A_kA k plus 2 times a square number, LaTeX: 25=\color{red}5\color{black}^2.

 

 

 

 

 

 

 

While it was originally thought that a single index could be used from the original example (see Figure 2), from observing the behavior in each of the subsets (Figures 3-7) it was discerned that second index, the n, would be needed.  In fact, it was realized that a double index would be needed to realize a method for generating a covering of the primitive Pythagorean Triples.

The double index method that encompasses the behaviors observed is as follows:

LaTeX: \begin{array}{ll} A_{kn} = (2k+1)((2k+1)+2(\color{red}n\color{black}))=(2k+1)(2k+2n+1)\\ B_{kn} = 4\left(\frac{\color{red}n\color{black}(\color{red}n\color{black}+1)}{2}\right)+4(\color{red}n\color{black})(k)=4\left(\frac{n(n+1)}{2}\right)+4kn\\ C_{kn}=A_{kn}+2(\color{red}n\color{black}^2)=(2k+1)(2k+2n+1)+2n^2 \end{array}

where LaTeX: k\in W LaTeX: n\in N, and LaTeX: 2k+1 and LaTeX: n are relatively prime. An algabraic proof can be used to show that every set of LaTeX: A_{kn},~B_{kn} and LaTeX: C_{kn} is a Pythagorean Triple.


Algebraic Proof

Pythag8.PNG

This shows that the LaTeX: BLaTeX: A, and LaTeX: C, which were defined using the triangular numbers, can be used to generate Pythagorean Triples.  Furthermore, it would be useful to show that this generator produces a covering of the Primitive Pythagorean Triples.  To prove this, the LaTeX: BLaTeX: A, and LaTeX: C will be related to Euclid’s formula, which does provide a covering of the Primitive Pythagorean Triples.

As mentioned previously, Euclid’s formula uses two positive, co-prime integers LaTeX: m_em e and LaTeX: n_en e, where imagen_e” src=”https://howardcc.instructure.com/equation_images/m_e%253En_e” alt=”LaTeX: m_e>n_e” data-equation-content=”m_e>n_e” data-mathml=” m e > n e “> and LaTeX: m_e-n_e is odd. Using LaTeX: m_eand LaTeX: n_e, Euclid’s formula defines LaTeX: ALaTeX: B, and LaTeX: C, as:

LaTeX: A_e=m_e^2-n_e^2~~~~~~~~~~~~~~~B_e=2m_en_e~~~~~~~~~~~~~~~C_e=m_e^2+n_e^2

The first step in relating the new method with Euclid’s formula is to determine if LaTeX: A_{kn},~B_{kn} and LaTeX: C_{kn} can be written in the same form as LaTeX: A_e,~B_e and LaTeX: C_e respectively.

Pythag9.PNG

It appears as though LaTeX: 2k+1+n and n, from the new method, are equal to the LaTeX: m_e and LaTeX: n_e from Euclid’s formula respectively. Proving that this is in true is the next step in relating the new method to Euclid’s formula and is done below.


Proof LaTeX: m_e=2k+1+1m e = 2 k + 1 + 1 and LaTeX: n_e=nn e = n

Suppose LaTeX: m_e=2k+1+n and LaTeX: n_e=n where LaTeX: k\in Wand n ∈ N.  First, it must be shown that LaTeX: m_e and LaTeX: n_e are positive integers.  From the supposition, LaTeX: m_e and LaTeX: n_eare the sums of the products of nonnegative integers. Thus, LaTeX: m_e and LaTeX: n_eare integers. The smallest possible values of LaTeX: k and LaTeX: n are 0 and 1 respectively. Let LaTeX: k=0and LaTeX: n=1. Then LaTeX: m_e=2(0)+1+(1)=1+1=2 and LaTeX: n_e=1. Since LaTeX: m_e and LaTeX: n_e are positive integers when LaTeX: k and LaTeX: n are their smallest values, LaTeX: m_eand LaTeX: n_e will be positive integers for any values of LaTeX: k and LaTeX: n.  It must also be shown that imagen_e” src=”https://howardcc.instructure.com/equation_images/m_e%253En_e” alt=”LaTeX: m_e>n_e” data-equation-content=”m_e>n_e” data-mathml=” m e > n e “>.  The smallest possible value of LaTeX: k is 0. Let LaTeX: k=0 Then LaTeX: 2k+1=2(0)+1=1. Since the smallest value of LaTeX: k results in a positive value for LaTeX: 2k+1, by the definition of odd, LaTeX: 2k+1 will always be a positive odd integer. Suppose imagen_e” src=”https://howardcc.instructure.com/equation_images/m_e%253En_e” alt=”LaTeX: m_e>n_e” data-equation-content=”m_e>n_e” data-mathml=” m e > n e “>. Then, imagen_e” src=”https://howardcc.instructure.com/equation_images/m_e%253En_e” alt=”LaTeX: m_e>n_e” data-equation-content=”m_e>n_e” data-mathml=” m e > n e “> so imagen” src=”https://howardcc.instructure.com/equation_images/2k%252B1%252Bn%253En” alt=”LaTeX: 2k+1+n>n” data-equation-content=”2k+1+n>n” data-mathml=” 2 k + 1 + n > n “>  and image0″ src=”https://howardcc.instructure.com/equation_images/2k%252B1%253E0″ alt=”LaTeX: 2k+1>0″ data-equation-content=”2k+1>0″ data-mathml=” 2 k + 1 > 0 “>Since LaTeX: 2k+1 is a positive odd integer it will always be greater than 0. Therefore, imagen_e” src=”https://howardcc.instructure.com/equation_images/m_e%253En_e” alt=”LaTeX: m_e>n_e” data-equation-content=”m_e>n_e” data-mathml=” m e > n e “>.

Further, it must also be shown that LaTeX: m_e-n_e is odd for LaTeX: m_e and LaTeX: n_e to have opposite parity. The smallest possible value of LaTeX: k is 0. Let LaTeX: k=0. Then LaTeX: 2k+1=2(0)+1=1. Since the smallest value of LaTeX: kk results in a positive value for LaTeX: 2k+1, by the definition of odd, LaTeX: 2k+1 will always be a positive odd integer. Thus, LaTeX: m_e-n_e=(2k+1+n)-n=2k+1 and so LaTeX: m_e-n_e is a positive odd integer.

This all shows that LaTeX: 2k+1+n and LaTeX: n from the new method are equal to the LaTeX: m_e and LaTeX: n_e from Euclid’s formula, respectively.

The final step in relating the new method to Euclid’s formula is to prove that the requirement that LaTeX: 2k+1 and LaTeX: n be relatively prime for the new method to generate primitive triples is equivalent to the requirement that LaTeX: m_e and LaTeX: n_ebe relatively prime for Euclid’s formula to generate primitive triples.


Proof that LaTeX: m_em e and LaTeX: n_en e being relatively prime is logically equivalent to LaTeX: k+1k + 1 and LaTeX: nn being relatively prime.

Let LaTeX: 2k+1 and LaTeX: n have a greatest common divisor LaTeX: x\in N. Then LaTeX: 2k+1=ax and LaTeX: n=bx where LaTeX: a\in NLaTeX: b\in N, and LaTeX: gcd(a,b)=1. Thus, LaTeX: m_e=2k+1+n=ax+bx=x(a+b)) and LaTeX: n_e=n=bx=x(b) Since LaTeX: a+b is the sum of natural numbers, LaTeX: (a+b)\in N. Therefore, LaTeX: m_e and LaTeX: n_e are products of LaTeX: x and some natural number, LaTeX: (a+b) and LaTeX: b respectively. In addition, because , LaTeX: gcd(a,b)=1,~gcd((a+b),b)=1

Pythag10.PNG

Thus, LaTeX: m_e and LaTeX: n_e are relatively prime if, and only if, LaTeX: 2k+1 and LaTeX: n are relatively prime.


Summary

The results in this paper answer the question of whether triangular numbers can be used to generate Primitive Pythagorean Triples.  It was shown that a double index was necessary for such a formula to generate Primitive Pythagorean Triples. It was also shown that this generating formula provides a covering of the Primitive Triples by relating the formula to that of Euclid. Further inquiry will investigate a sieve for separating the triples that are generated.

 

Contact:  rn28456@umbc.edu, ksutton1@umbc.edu

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