Pythagorean Triples with a Fixed Difference between a Leg and the Hypotenuse

Transcription

1 Pythagorean Triples with a Fixed Difference between a Leg and the Hypotenuse Anna Little Advisor: Michael Spivey April, 005 Abstract This project examines sets of Pythagorean triples with a fixed difference between one of the legs and the hypotenuse. A Pythagorean triple is an ordered triple of positive integers (x, y, z) such that x + y = z. This paper proposes and proves a formula that will generate every triple of a difference, that is, every triple where either z x = or z y =. The proof of this result utilizes the general formula for all primitive Pythagorean triples as well as the prime factorization of the integer. 1 Introduction Pythagorean triples are a very ancient area in mathematics and throughout the centuries numerous mathematicians have deepened our understanding of their characteristics. Sierpinsi provides an excellent summary on some of their most well nown properties in [6]. By far the most significant discovery in this area was made by Euclid (c. 350 B.C.), who found a formula that characterized every primitive Pythagorean triple []. One of the first nown references to Pythagorean triples of a fixed difference is found in the writings of Plato. Plato notes that (4n, 4n 1, 4n + 1) yields an infinite set of Pythagorean triples of difference as n ranges over the positive integers [1]. Building on these foundations, this project sought to find a formula that would yield every Pythagorean triple of an arbitrary positive difference. Numerous formulas for specific values were found and analyzed until a correct formula was obtained. As noted, the proof of the formula utilizes Euclid s classification of primitive Pythagorean triples and well-nown number theoretic properties. Although similar results have been published [4], the formula and its proof were arrived at independently. Formula for Pythagorean Triples of a Difference We have the following theorem as our main result: Theorem 1. Define where n0 p n1 1 pn...pnr r = n 0 p n 1 1 p n...p nr r, (.1) is the prime factorization of. Then every triple of difference is given by: ( (m + mod ), (m )(m + ), (m )(m + ) + ) as m runs through all positive integers greater than. Now by Euclid s general formula for all primitive Pythagorean triples [], any triple can be written as (l(mn), l(m n ), l(m + n )), where l is some positive integer and m > n > 0 are two relatively prime positive integers of opposite parity. Throughout this paper x will be used to designate the leg that is a multiple of mn, y will be used to designate the leg that is a multiple of m n, and z will designate the hypotenuse. Thus (x, y, z) and (y, x, z) are simply two different notations for the same triple and will be used interchangeably. 1

2 We will discuss briefly the outline of the proof of this theorem, and then we will proceed to the actual proof. Let S be the set of all triples of difference, that is, the set of every triple where the hypotenuse is separated by from a leg. Let G be the set of all triples generated by the above formula as m runs over all the positive integers greater than. We will show that these sets are equal, that is, that S G and that G S. To show that S G, we will choose an arbitrary triple (x, y, z) of difference and show that it is infact generated by the formula given in Theorem 1. In doing this we will examine two cases: is even and is odd. When is even, we will have two subcases: z x = and z y =. When is odd, however, we will have only z x =. After showing that in every case the triple (x, y, z) is generated by the formula given in Theorem 1, we conclude that (x, y, z) G and thus that S G. To show that G S, we pic an arbitrary element (q, r, s) that is generated by the formula given in Theorem 1, and we show that it is infact a Pythagorean triple of difference. To do this, we have to verify that q, r, and s are all positive integers, that q + r = s, and that either s q = or s r =. After showing that (q, r, s) is indeed a Pythagorean triple of difference, we conlcude that (q, r, s) S and therefore that G S. We will then have shown that the sets are equal and thus that the formula given in Theorem 1 produces every triple of differnce and nothing else. We will now begin the actual proof. First of all, we show that S G. Let be an arbitrary positive integer and let (x, y, z) S. We will loo at the cases where is even and where is odd, and we will show that in both cases (x, y, z) G. Case 1. Let be an even integer. By the general formula for primitive Pythagorean triples, (x, y, z) is of the form: x = l(mn) y = l(m n ) (.) z = l(m + n ), where l is some positive integer and m > n > 0 are two relatively prime positive integers of opposite parity. Now since this triple has a difference, either z x = or z y =. We will examine both of these two subcases. Subcase 1: Suppose that z y =. Then l(m + n ) l(m n ) =, which yields = ln. Thus n =, and substituting this expression for n into equation., we get: l x = m l y = lm (.3) z = lm +. We will use the following lemma to show (x, y, z) G : Lemma 1. If (x, y, z) is a triple where z y =, and thus has the form given above, it can be written in the form x = m 1 l1 y = l 1 m 1 (.4) z = l 1 m 1 +, for some integers l 1 and m 1, where l 1 is the smallest positive integer such that l 1 is a perfect square and m 1 = md, where d = l. l1 Proof. Let q = l 1, where l 1 is the smallest positive integer such that l 1 is a perfect square. Let p = l, which must be an integer since x is an integer. We will show that q divides p. Now q = l 1 = n 0p n pnr r l 1, where n0 p n1 1...pnr r is the prime factorization of. Now let p o1, p o,..., p oj be the primes

3 with odd exponents and let p e1, p e,..., p es be the primes with even exponents. Since l 1 is the smallest integer such that q Z, l 1 = p o1 p o...p oj. Now let n o1, n o,..., n oj be the odd exponents and let n e1, n e,..., n es be the even exponents. Then q = p n e1 e1pe n e...pes n espo1 n o1 +1 p n o o +1 n...p oj +1 oj. (.5) Now consider p = l. We have: p = p n e1 e1pe n e...pes n espo1 n o1po n o...poj n oj l. (.6) Thus in order for p to be an integer, l must contain at least one copy of each of the primes with odd l exponents. Thus p o1 p o...p oj l. Let r = p o1 p o...p oj. Thus: p = p e1 n e1pe n e...pes n espo1 n o1 +1 p o n o +1...p oj n oj +1 r. (.7) Now since every prime exponent is even, every exponent in the prime factorization of r must also be even, or p won t be an integer. Thus r must be a perfect square, and r = d, for some positive integer d. Thus: p q = pe1 Thus p = qd, and q p. Now let m 1 = md. Then: ne1p e ne...p es nesp o1 no1 +1 p o no +1...p noj +1 oj r pe1 ne1p e ne...p es nesp o1 no1 +1 p o no +1...p noj +1 oj = r = d. m 1 l1 = md l 1 = m l 1 r = m l = x (since l = p o1 p o...p oj r = l 1 r) l 1 m 1 = l 1(md) = l 1m r = lm = y (.8) l 1 m 1 + = l 1(md) + = l 1m r + = lm + = z. Thus we have shown that the triple (x, y, z) we selected, where z y =, can be written in the form: x = m 1 l1 y = l 1 m 1 (.9) z = l 1 m 1 +, for some integers l 1 and m 1, where l 1 is the smallest positive integer such that l 1 is a perfect square. Now we need to show that this triple is in the set G. We will do this by showing that there exists some integer m > such that: x = (m + mod ) y = (m )(m + ) (.10) z = (m )(m + ) +. Before we can show this, however, we must prove an additional lemma, which holds regardless of the parity of. 3

4 Lemma. If l 1 is the smallest positive integer such that l 1 is a perfect square and is as defined in Theorem 1, then l 1 =. Proof. Let n0 p n1 1 pn...pnr r be the prime factorization of. Now let p o1, p o,..., p oj be the primes with odd exponents and let p e1, p e,..., p es be the primes with even exponents; also, let n e1, n e,..., n es be the even exponents and let n o1, n o,..., n oj be the odd exponents. Then: = n 0p n 1 1 pn...pnr r ( n 0 p n 1 1 p n...p nr r ) = p e1 ne1p e ne...p es nesp o1 no1p o no...p oj noj ne 1 (p ne e 1 p nes e...p e s ) (p o no 1 no 1 p o...p o noj j ) = p e1 ne 1p e ne...p es nesp o1 no1p o no...p noj oj p ne1 e 1 p ne e...p nes e s p no1 1 o 1 p no 1 noj o...p 1 o j = p o1 p o...p oj = l 1. (Since l 1 is the smallest integer such that l 1 is a perfect square, it contains exactly one copy of each prime that has an odd exponent in the prime factorization of.) Now we will proceed to show that the integer m 1, as described in Lemma 1, is the integer that will generate the triple (x, y, z) when inserted into our formula for G, as given in Theorem 1. Since y > 0 and y = l 1 m 1, we have: l 1 m 1 > 0 l 1 m 1 > m 1 > (using the substitution from Lemma ) 4m 1 > (.11) m 1 > m 1 > m 1 >. Thus since m 1 is an integer and m 1 >, m 1 is in the domain of the formula given in Theorem 1. Now 4

5 since is even, we have: x = m 1 l1 = m 1 = m 1 = (m 1 + mod ) y = l 1 m 1 = m 1 4 = (m 1 4 ) (.1) = (m 1 )(m 1 + ) = (m 1 )(m 1 + ) z = l 1 m 1 + = l 1 m 1 + = (m 1 )(m 1 + ) +. Note that since is even, is also clearly even, since has at least two factors of in it. Thus = =, which was used in the above derivation. Thus we have shown that when is even and when z y =, (x, y, z) G because the integer m 1 produces it. Subcase : Now suppose that z x = ( still even). We will mae use of the following lemma to show that (x, y, z) G. Note, however, that this lemma does not require to be even. Lemma 3. If (x, y, z) is a triple such that z x =, then there exist integers m and l such that x = m (m l l ) y = m l (.13) z = m (m l l ) +, where l is the smallest positive integer such that l is a perfect square. Proof. Let (x, y, z) be a triple where z x =. By the general formula for primitive Pythagorean triples, x = l(mn) y = l(m n ) (.14) z = l(m + n ), where l is some positive integer and m > n > 0 are two relatively prime positive integers of opposite parity. Since z x =, we have l(m + n ) l(mn) =, which yields n = m. Substituting this expression l 5

6 for n into equation.14, we get: x = m(ml l) y = m l (.15) z = m(ml l) +. Now we need to show that there exists some integer m such that: x = m (m l l ) y = m l (.16) z = m (m l l ) +, where l is the smallest positive integer such that l is a perfect square. Let q = l and let p = l (note that p must be an integer for x, y, and z to be integers). Let a0 p a1 1 pa...par r be the prime factorization of. Also, let p o1, p o,..., p oj be the primes with odd exponents and let p e1, p e,..., p es be the primes with even exponents. Now let a o1, a o,..., a oj be the odd exponents and let a e1, a e,..., a es be the even exponents. We will show that q divides p. Now, q = l = p a0 0 pa1 1 pa...par r l = p a e1 e1pe a e...pes a espo1 a o1po a o...poj a oj l. (.17) Since l is the smallest integer that will mae every exponent under the radical even, l = p o1 p o...p oj, and Now, p = l = l = p e1 a e1pe a e...pes a espo1 a o1 +1 p o a o +1...p oj a oj +1. (.18) p a0 0 pa1 1 pa...par r l = p a e1 e1pe a e...pes a espo1 a o1po a o...poj a oj l. (.19) Now in order for p to be an integer, l must contain as factors at least one copy of every prime with an odd exponent in the prime factorization of ; thus l l, so l = l r, for some integer r. The integer r must also contain only even exponents in its prime factorization, or p will again not be an integer. Thus r = d, for some integer d. So we have: l = p e1 a e1pe a e...pes a espo1 a o1 +1 p o a o +1...p oj a oj +1 r, (.0) and therefore p q = l l = p a e1 e1pe a e...pes a espo1 a o1 +1 p a o o +1 a...p oj +1 oj r = r = d. (.1) p a e1 e1pe a e...pes a espo1 a o1 +1 p a o o +1 a...p oj +1 oj Thus q p. Now let m = md. Then: m (m l l ) = md(mdl l ) = m rl md l = m l m l r = m l m l = m(ml l) = x (.) 6

7 m l = md l = m l r = m l = y m (m l l ) + = md(mdl l ) + = m(ml l) + = z. (.3) Thus we have shown that for any triple (x, y, z) where z x =, there exist integers m and l such that x = m (m l l ) y = m l (.4) z = m (m l l ) +, where l is the smallest positive integer such that l is a perfect square. Now we are trying to show that when is even and z x =, our triple (x, y, z) G. In Lemma 3, we established the general form that any triple (x, y, z) where z x = must have. We now state an additional lemma that provides a simplification of this form when is required to be even. Lemma 4. Suppose (x, y, z) is a triple such that z x = for some even. Then there exists some integer m 1 such that x = l 1 m 1 y = m 1 l1 (.5) z = l 1 m 1 +, where l 1 is the smallest integer such that l 1 is a perfect square. Proof. Let (x, y, z) be a triple such that z x = for some even. Now by Lemma 3, (x, y, z) has the following form: x = m (m l l ) y = m l (.6) z = m (m l l ) +, for some integers m and l, where l is the smallest positive integer such that l is a perfect square. Now is even, and it has either an even or an odd number of s in its prime factorization. We will show that in either case, an integer m 1 can be found that will allow us to write the triple in the form given in Lemma 4. Case 1: Suppose has an odd number of s in its prime factorization. Then = p e1 n e1pe n e...pes n es no1 p o n o...poj n oj, where, po,..., p oj are the prime factors with odd exponents and p e1, p e,..., p es are the prime factors with even exponents. Recall that since l 1 is the smallest integer such that l 1 is a perfect square, l 1 = p o...p oj (since contains an odd number of s, there is no need for an additional factor of to mae l 1 a perfect square; thus is not a factor of l 1 ). However, l must have a factor of in order for l to be a perfect square; namely, l = p o...p oj. Therefore, l 1 = l, that is l = l 1. Thus: l 1 = p e1 n e1 p e n e...p es n es no1 +1 p o n o +1...p oj n oj +1, (.7) 7

8 which implies that l1 = p e1 ne1 p e ne...p es nes no1 p o no...p oj noj ne1 ne nes p e1 p e...p es no1+1 no p +1 noj +1 o...p oj = p e1 ne1 p e ne...p es nes no1 1 p o no 1...p oj noj 1. (.8) Now since n e1, n e,..., n es and n o1 1, n o 1,..., n oj 1 are all even, nonnegative integers, ne1, ne nes..., and no1 1, no 1,..., noj 1 are all nonnegative integers, and thus the above product is an integer. Namely, is an integer. l1 Now let m 1 = m form given in Lemma 4. We have: l1. We will show that this is the integer we need to write the triple (x, y, z) in the x = m (m l l ) = ( m1 + l 1 )( l1m1 + l1 l 1 l 1 ) = (m 1 + l1 )(l 1 m 1 + l1 l1 l 1 ) = (m 1 + l1 )(l 1 m 1 + l 1 l 1 ) = l 1 m 1 + m 1 l1 = l 1 m 1 + m 1 ( l 1 m 1 l1 + m 1 l1 + l 1 l 1 l1 + l 1 ) = l 1 m 1 + m 1 ( l 1 l 1 ) = l 1 m 1 (.9) y = m l = ( m1 + l 1 )( l 1 ) = m 1 l1 z = m (m l l ) + = l 1 m 1 + = l 1 m 1 +, since we just showed that x = m (m l l ) = l 1 m 1. Thus when there are an odd number of s in the prime factorization of, a triple (x, y, z) where z x = can be written in the form given in Lemma 4. Now we proceed to the second case in the proof of Lemma 4. Case. Suppose has an even number of s in its prime factorization. We will show that m 1 = m l1 is the integer that will allow us to write (x, y, z) in the desired form. First of all, however, we must show that l1 is indeed an integer. Since has an even number of s in its prime factorization, = ne1 n p e...pes n espo1 n o1po n o...poj n oj e, where p o1, p o,..., p oj are the prime factors with odd exponents and, p e,..., p es are the prime factors with 8

9 even exponents. Now since contains an even number of s, l 1 must contain a factor of in order for l 1 to be a perfect square. However, l does not need a factor of in order for l to be a perfect square, and thus l 1 = l. So we get l 1 = p o1 p o...p oj and l 1 = ne1 + p e n e...pes n espo1 n o1 +1 p o n o +1...p oj n oj +1. Thus: l1 = ne1 p e ne...p e s nes p o 1 no1 p o no...p o j noj ne1+ p e ne...pe s nes po 1 no1+1 p o no+1 noj +1...p o j = ne1 p e ne...p e s nes p o 1 no1 p o no...p o j noj ne1+ ne nes no1 p e...p e s p +1 no o 1 p +1 o...p o j noj +1 = ne1 ne nes no1 p e...p e s p 1 no o 1 p 1 noj o...p 1 o j. Now n e1, n e,..., n es and n o1 1, n o 1,..., n oj 1 are all even; thus all the exponents are integers. Note that since is even, n e1, and thus n e1 0. Thus we now that all of the exponents are in fact nonnegative integers, and thus the above product is an integer; namely, is an integer. Now let m 1 = m l1. Then x = m (m l l ) = (m 1 + l1 )( l1m1 + l1 l 1 l 1 ) = (m 1 + l1 )( l1m1 + l 1 l 1 ) = l 1 m 1 + m 1 l1 l1 m 1 l1 + m 1 l1 + = l 1 m 1 + m 1 ( l 1 l 1 + l 1 ) = l 1 m 1 y = m l = (m 1 + l l1 )( 1 ) = (m 1 + l l1 )( 1 ) (.30) = m 1 l1 + = m 1 l1 z = m (m l l ) + = l 1 m 1 + = l 1 m 1 +, since we already showed in our analysis of x that m (m l l ) = l 1 m 1. Thus we have shown that when (x, y, z) is a triple where z x = for some even, there exists some 9

10 integer m 1 such that the triple can be written as: x = l 1 m 1 y = m 1 l1 (.31) z = l 1 m 1 +, where l 1 is the smallest integer such that l 1 is a perfect square. We are trying to show that when (x, y, z) is a triple where z x = for some even, (x, y, z) G. Having shown in Lemma 4 that (x, y, z) has the above form, we can now use a derivation almost identical to the one in.1 to show that (x, y, z) G ; namely, we will show that when the integer m 1 is inserted into the formula given in Theorem 1, the triple (x, y, z) is produced. Note that since x > 0 and x = l 1 m 1, we have l 1m 1 > 0, and an identical argument to the one given in.11 shows that m 1 >. Thus m 1 is in the domain of the formula given in Theorem 1, and we have: x = l 1 m 1 = m 1 4 = (m 1 4 ) = (m 1 )(m 1 + ) = (m 1 )(m 1 + ) y = m 1 l1 = m 1 (.3) = m 1 = m 1 = (m 1 + mod ) z = l 1 m 1 + = l 1 m 1 + = (m 1 )(m 1 + ) +. Thus we see that when the integer m 1 is inserted into the formula given in Theorem 1, the triple (y, x, z) is produced. Since we are considering the triple (y, x, z) to be the same triple as (x, y, z), (x, y, z) G. In the above derivation we once again made use of the fact that l 1 =, which we proved in Lemma. Thus we have shown that when is even, (x, y, z) G for the subcases z y = and z x =. Since these are the only possible subcases, we have shown that when is even, (x, y, z) G. We now proceed to Case, where is odd. Case : Suppose that is an odd integer. Now by the general formula for primitive Pythagorean triples, 10

11 we now that (x, y, z) has the following form: x = l(mn) y = l(m n ) (.33) z = l(m + n ), where l is some positive integer and m > n > 0 are two relatively prime positive integers of opposite parity. Now since we are supposing this triple has difference, either z x = or z y =. Observe that z y = l(m + n ) l(m n ) = ln ; thus z y is even. Since is odd, z y, and we must have z x =. Thus for odd, we do not have two subcases lie we did for even. Now by Lemma 3 (since Lemma 3 holds for both even and odd ), the triple (x, y, z) can be written in the following form for some integer m : x = m (m l l ) y = m l (.34) z = m (m l l ) +, where l is the smallest positive integer such that l is a perfect square. We need to show that this triple is in the set G. We will proceed by showing that the triple (y, x, z) G, which clearly implies that (x, y, z) G. To show (y, x, z) G, we must show that there exists some integer m 1 greater than such that: y = (m 1 + mod ) x = (m 1 )(m 1 + ) (.35) z = (m 1 )(m 1 + ) +. Recall that l 1 is the smallest positive integer such that l 1 is a perfect square and that l is the smallest integer such that l is a perfect square. Thus l contains one copy of every prime factor of that has an odd exponent, while l 1 contains one copy of every prime factor of that has an odd exponent. Since is not a factor of, l does not contain a factor of, but since is a factor of and has an odd exponent in the prime factorization of, l 1 does have a factor of. However, all the other factors of l 1 and l are clearly the same. Thus we have l = l1. Recall that l 1 =, as shown in Lemma, which holds regardless of the parity of. So: l = l 1 = =. (.36) Now let m 1 = m +1. We will show that m 1 is the integer that will produce the triple (y, x, z) when inserted into the formula given in Theorem 1. First of all, however, we must show that m 1 is in the domain of the formula given in Theorem 1; namely, we must verify that m 1 is an integer and that m 1 >. Now since is odd, there is only one in the prime fctorization of, that is = 1 p n1 1 pn...pnr r, where, p 1, p,..., p r are the distinct prime factors of. Thus = 1 p n 1 1 p n...p nr r, which shows that has no factors of. m 1 = m +1 is also an integer. Thus is odd, maing + 1 even and +1 an integer. Therefore, 11

12 Note also that since x > 0 and x = m (m l l ), we must have m l l > 0. Thus: (m )l > l (m ) > m > m 1 > +1 (.37) m 1 > 1 m 1 > 1 (since is odd, is odd, and thus = ). Thus m 1 is an integer greater than and is thus in the domain of the formula given in Theorem 1. Since m = m , we get: y = m l = m = m = (m )( ) = (m )( ) = m = (m 1 + 1) = (m 1 + mod ) x = m (m l l ) = (m )[(m ) ] (.38) = (m )( m1 + + ) = (m )( m1 + ) = m 1 + m 1 m 1 + m 1 = m 1 + m 1 + = (m 1 + m ) + = (m 1 + m 1 ( +1 1 ) ( 1 4 )) + m 1 = (m 1 m 1 ( 1 ) + m 1 ( +1 ) ( 1 )( +1 )) + 1

13 Now since is odd, is odd, and thus we have that = 1 and = +1. Substituting this we get: x = (m 1 m 1 + m 1 ) (.39) = (m 1 )(m 1 + ). Liewise, we see that: z = x + = (m 1 )(m 1 + ) +. (.40) Thus when (x, y, z) is a triple where z x = for some odd integer, (y, x, z) is clearly an element of G because it is the triple that is generated when the integer m 1 is inserted into the formula given in Theorem 1. Since, again, we are considering (y, x, z) to be the same triple as (x, y, z), we have shown that (x, y, z) G. Thus we have shown that for an arbitrary triple (x, y, z) of difference, that is for an arbitrary element of S, (x, y, z) G both when is even and when is odd. Therefore we have shown that S G. For even, we examined two subcases: z y = and z x =, while for odd we had only z x =. To conclude the proof of our main theorem, we must now show that G S. Let (q, r, s) be an arbitrary element of G and let m and be the integers required by Theorem 1 to produce (q, r, s). In order for (q, r, s) S, we need to verify that (q, r, s) is a Pythagorean triple of difference. Note that if (q, r, s) is a Pythagorean triple, then it is clearly a Pythagorean triple with a difference of. For since (q, r, s) has the following form: q = (m + mod ) r = (m )(m + ) (.41) s = (m )(m + ) +, clearly s r =. Therefore, all that needs to be shown is that (q, r, s) is a Pythagorean triple, namely that q, r, and s are positive integers and that q + r = s. Now since m >, clearly q, r, and s must all be positive. Thus if and are integers, then q, r, and s must in fact be positive integers. It is clear from Lemma that is an integer; thus we will simply show that is an integer. We will loo at the even and odd cases. Case 1: Suppose is even and let n0 p n1 1...pnr r be the prime factorization of. Then: = n 0 +1 = n0 n 0 +1 n0 p n1 1...pnr r p n 1 1 p n...p nr r p n1 n 1 n n 1 p...p Now since every exponent in the above expression is clearly nonnegative, nr nr r. is an integer. (.4) Case : Suppose is odd and that = p n1 1 pn...pnr r is the prime factorization of. Then: = p n1 1 pn...pnr r 1 p n...p nr r 1 p n 1 = p n 1 1 pn...pnr r 0 p n 1 1 p n...p nr r (.43) n n 1 p = p n1 n 1...p Once again, since every exponent is clearly a nonnegative integer, nr nr r is also an integer. Thus we have verified that q, r, and s are indeed positive integers. Now we will show that q + r = s. Once again, we will loo at two cases. 13

14 Case 1: Suppose that is even. Then since is even, is also even because it clearly contains a factor of. Thus = =. Also, mod = 0. Thus our formulas for q, r, and s can be simplified as follows: q = (m) r = (m )(m + ) (.44) Thus: s = (m )(m + ) +. q + r = (m ) + [ (m )(m + )] = (m ) + [ (m 4 )] = (m ) + ( m ) = 4m + 4 m 4 4 m + 4 = 4 m m + 4 = ( m + ) (.45) = ( m + ) = [ (m 4 ) + ] = [ (m )(m + ) + ] = s Case : Suppose that is odd. Then is also odd since it does not contain a factor of. Thus = 1 and = +1. Also, mod = 1. Thus our formulas for q, r, and s simplify as follows: q = (m + 1) r = (m 1 )(m + +1 ) (.46) Thus: s = (m 1 )(m + +1 ) +. q + r = [ (m + 1)] + [ (m 1 )(m + +1 )] = (4m + 4m + 1) [(m 1 )(m + +1 )] = 4 m + 4 m [(m 1 4 )(m + +1 )] (.47) = 4 m + 4 m [(m 1 4 )(m + +1 )] 14

15 = 4 (m + m ) [(m 1 )(m + +1 )] = 4 (m 1 )(m + +1 ) [(m 1 )(m + +1 )] = 4 4 [(m 1 )(m + +1 )] + 4 (m 1 )(m + +1 ) + (.48) = [ (m 1 )(m + +1 ) + ] = s So we see that (q, r, s), both for even and odd, is indeed a Pythagorean triple with a difference of, and so (q, r, s) S. Thus we have shown that G S. Finally, since S G and G S, G = S, which concludes the proof of our main theorem. Thus the formula given in Theorem 1 produces every triple of difference for any positive integer. 3 Alternate Formula for Pythagorean Triples of Difference Having proved that G = S, we can now prove some additional properties about S using this equivalence. In particular, we have the following result: Theorem. The set of all triples of difference can be generated by the union of disjoint, infinite subsets, where the following property holds for each of these subsets: d Leg(n) = (Leg1(n)), (3.1) dn where (Leg1(n), Leg(n), Leg(n) + ) denote the generating formulas of the subset. In particular, for even: S = 1 i=0 {(n + i, n + 4i n + i, (3.) n + 4i n + i + ) : n Z, n > 1 i }. Similarly, for odd: S = 1 i=0 {(n + i +, n + 4i n + n + i + i + 1 4, (3.3) n + 4i n + n + i + i ) : n Z, n > 1 1 i }. The proof of this theorem is a clear application of Theorem 1. Once again, we will examine both the even and odd cases. Proof. Let be some positive integer. Now by Theorem 1, every triple of difference is generated by the following formula as m ranges over the positive integers greater than : ( (m + mod ), (m )(m + ), (m )(m + ) + ). (3.4) Case 1: Suppose is even. Then our generating formula simplifies to: ( (m), (m )(m + ), (m )(m + ) + ). (3.5) Thus as m ranges over the positive integers greater than every triple of difference is produced. It is easy to see that every distinct positive integer produces a unique triple when inserted into the formula. Now 15

16 by the division algorithm [], we can partition all the positive integers greater than into their residue classes mod. If we were partitioning all of the positive integers into their residue classes mod, the ith residue class would be given by n + i, where n is any nonnegative integer. However, to eliminate producing integers in the ith residue class that are smaller than, we must have n + i >, which happens exactly when n > 1 i. Thus the set {n + i : n Z, n > 1 i } is exactly the ith residue class mod of the positive integers greater than. The restriction on n causes n to start at 0 or 1 as necessary, so that an integer smaller than is never produced. Now running m over this particular residue class will produce a subset of S. In order to only produce the triples generated by the ith residue class, we must substitute n + i for m in the formula given in Theorem 1 and allow n to range over the integers greater than 1. After maing this substitution into our formula, we find that i (n + i, n + 4i n + i, n + 4i n + i + ) (3.6) gives every triple generated by the ith residue class of the domain of Theorem 1 as n ranges over the integers greater than 1 i. Now since every triple produced by our original formula is unique and the residue classes are disjoint, the subsets produced by these residue classes will clearly be disjoint. Also, if we tae the union of the triples generated by every residue class we will have the triples generated by every positive integer greater than, that is, we will have the set of all triples of difference. Thus we find that, for even: S = 1 i=0 {(n + i, n + 4i n + i, n + 4i n + i + ) : n Z, n > 1 i }, (3.7) as stated in Theorem. Thus we have rewritten S as the union of disjoint, infinite subsets. Notice also that, allowing Leg1(n) = n + i and Leg(n) = n + 4i n + i to denote the generating formulas for the two legs in the ith residue class, the following property holds for each of the subsets in the above union: d dn Leg(n) = d dn (n + 4i n + i ) = 4n + 4i = (n + i ) (3.8) = (Leg1(n)). Thus we have seen that Theorem holds for even. Case : Let be an odd positive integer. Then our generating formula for S simplifies as follows: ( (m + 1), 1 (m )(m ), 1 (m )(m ) + ). Thus as m ranges over the positive integers greater than, every triple of difference is produced. Once again, we partition the positive integers greater than into their residue classes mod. We 16

17 must ensure that only those integers greater than are produced, namely, we must have: n + i > n + i > 1 n + i > 1 (3.9) n > 1 i n > 1 1 i. Thus n+i, as n ranges over the integers greater than 1 1 i, produces the ith residue class mod of the integers greater than. Therefore, substituting n + i for m in the Theorem 1 formula and allowing n to range over the integers greater than 1 1 i produces the triples generated by the ith residue class. Maing this substitution and simplifying, we get: (n+ i +, n + 4i i n+ n+ + i +1 4, n + 4i i n+ n+ + i ) (3.10) as the generating fomula for the triples produced by the ith residue class. Thus, once again, since every triple produced by the Theorem 1 formula is unique and the residue classes are clearly disjoint, the subsets produced by the residue classes are disjoint. If we tae the union of these subsets, we will get every triple generated by the Theorem 1 formula, that is, every triple of a difference. Thus we can rewrite the set of all triples of difference as: S = 1 i=0 {(n + i +, n + 4i n + n + i + i + 1 4, (3.11) n + 4i n + n + i + i ) : n Z, n > 1 1 i }. Once again allowing Leg1(n) = n+ i + and Leg(n) = n + 4i i n+ n+ + i to be the generating formulas for the legs of the triples produced by the ith residue class, we find that: d dn Leg(n) = d dn (n + 4i n + n + i + i ) = (4n + 4i + ) = (n + i + ) (3.1) = (Leg1(n)). Thus for odd we can very similarly write S as the union of disjoint, infinite subsets where the property d dnleg(n) = (Leg1(n)) holds for the generating formulas of the legs of each subset. It is interesting to note that this derivative property in general does not hold for the generating formulas of the legs of the larger set S. 4 Conclusion In summary, we have found and proved two separate but connected formulas that will generate every Pythagorean triple of a fixed difference. First of all, we found that when when we define = n 0 p n 1 1 p n 17...p nr r, (4.1)

18 where n0 p n1 1 pn...pnr r is the prime factorization of, then every triple of a difference is given by: ( (m + mod ), (m )(m + ), (m )(m + ) + ) as m runs through the positive integers greater than. To prove this result, we let (x, y, z) be an arbitrary Pythagorean triple with a difference of and examined both the even and odd cases. For even, we had two subcases: z x = and z y =. For odd, we had only z x =. We showed that in every case the triple would be generated by our formula. We also showed that everything generated by the above formula is indeed a Pythagorean triple with difference. Having completed the proof of our main theorem, we then used the theorem to find an alternate formula that also generates every triple of difference. Namely, we found that, for even, every triple of a difference is given by the following: 1 i=0 {(n + i, n + 4i n + i, n + 4i n + i + ) : n Z, n > 1 i }, (4.) and that for odd, every triple of difference is given by: 1 i=0 {(n + i +, n + 4i n + n + i + i + 1 4, (4.3) n + 4i n + n + i + i ) : n Z, n > 1 1 i }. For each subset in the above unions we noted that the following property holds for the generating formulas of the legs of the triples: d Leg(n) = (Leg1(n)). (4.4) dn A better understanding of why the above property holds for the given subsets but not for the entire set of Pythagorean triples of difference, as well as an investigation into other unions that yield the entire set, are areas of future research. Also, the formula in Theorem 1 appears to produce the negative Pythagorean triples. The proof of the theorem, however, made use of Euclid s classification, which only holds for positive triples. Thus another area of future research is investigating whether the formula given in Theorem 1 and its proof can be extended to include all negative Pythagorean triples. 18

19 References [1] Apostol, Tom M. Introduction to Analytic Number Theory. Springer-Verlag, Inc., New Yor, [] Burton, David M. Elementary Number Theory. McGraw-Hill, New Yor, 00. [3] Konvalina, John. A characterization of the Pythagorean triples. A collection of manuscripts related to the Fibonacci sequence, pp , Fibonacci Assoc., Santa Clara, CA, [4] McCullough, Darryl. Height and Excess of Pythagorean Triples. Mathematics Magazine. Vol. 78(005), no. 1, pp [5] Wade, Peter W.; Wade, William R. Recursions that produce Pythagorean triples. College Math Journal. Vol. 31(000), no., pp [6] Sierpinsi, Waclaw. Pythagorean Triples. Graduate School of Science, Yeshiva University. New Yor, 196. [7] Turnbull, Herbert Westren. The Great Mathematicians. Barnes and Noble Boos, New Yor,