NEW GENERAL FORMULAS FOR PYTHAGOREAN TRIPLES

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NEW GENERAL FORMULAS FOR PYTHAGOREAN TRIPLES EDUARDO CALOOY ROQUE Abstract. Ths paper shows the General Formulas for Pythagorean trples that were derved from the dfferences of the sdes of a rght trangle. In addton, as computatonal proof, tables were made wth a C++ scrpt showng prmtve Pythagorean trples and ncluded as text fles and screenshots. Furthermore, to enable readers to check and verfy them, the C++ scrpt whch wll nteractvely generate tables of Pythagorean trples from the computer console command lne s attached. It can be run n Clng and ROOT C/C++ nterpreters or compled. 1. Introducton I beleve t would be best to proceed mmedately to the problem at hand,.e. to fnd general formulas for Pythagorean trples, rather than dscuss many thngs and go n crcles. So gong straght to the pont we wll derve and prove the relevant theorem then derve from t the formulas gven by the Greeks, Plato and Pythagoras. Then we wll show the general formulas that generates Pythagorean trples. At ths pont, the pattern wll manfest. Next we wll provde tables from a C++ scrpt to demonstrate valdty. In concluson, attached the C++ scrpt nstead of typesettng verbatm so that the document wll not look unnecessarly large. 2. Pythagorean Trples Defnton. If a, b, c N and a < b < c or b < a < c then a Pythagorean trple s a trple of natural numbers such that a 2 + b 2 = c 2. It s sad to be prmtve f (a, b, c) s parwse relatvely prme and the partes of a and b are always oppostes whle c s always odd. Theorem 1. If (a, b, c) s a Pythagorean trple then there exsts α, β, γ Z, k, n N where α = b a, β = c a, γ = c b, β = α + γ, k = n, n = 1, 2, 3,... such that (a γ) 2 = 2γβ γβ = 2k 2 a = γ + 2k b = β + 2k c = γ + β + 2k Date: November 21, 2012. 1

2 EDUARDO CALOOY ROQUE Proof. a 2 + b 2 = c 2 a 2 + (a + α) 2 = (a + β) 2 a 2 + (a 2 + 2aα + α 2 ) = a 2 + 2aβ + β 2 a 2 + 2a(α β) = β 2 α 2 a 2 2a(β α) = (β + α)(β α) a 2 2γa = [(α + γ) + α]γ (a γ) 2 = γ(2α + γ) + γ 2 (a γ) 2 = 2γ(α + γ) (a γ) 2 = 2γβ It s evdent that (a γ) 2 = 2γβ has even party and hence of the form 4k 2 where k, n N, k = n, n = 1, 2, 3,... Thus (a γ) 2 = 4k 2 and γβ = 2k 2. We now see that a = γ + 2k. Now β α = (c a) (b a) = c b. Ths s γ thus α + γ = β. From b a = α we get b = a + α = β + 2k and snce c b = γ we also get c = b + γ = γ + β + 2k. Theorem 2. If (a, b, c) s a Pythagorean trple then there exsts α, β, γ Z, k, n N where α = b a, β = c a, γ = c b, β = α + γ, (a γ) 2 = 2γβ, γβ = 2k 2, k = n, n = 1, 2, 3,... such that γ = 1, β = 2k 2, a = 2k + 1 b = 2k(k + 1) c = 2k 2 + 2k + 1 γ = 2, β = k 2, a = 2(k + 1) b = k(k + 2) c = k 2 + 2k + 2 Proof. From Theorem1, we have β = 2k2 γ and so Pythagorean trples are generated by a = γ + 2k, b = 2k2 γ + 2k, c = γ + 2k2 γ + 2k It s seen then that ntegral values can be obtaned for γ = 1, 2. If γ = 1 then β = 2k 2 and a = 2k + 1, b = 2k(k + 1), c = 2k 2 + 2k + 1. If γ = 2 then β = k 2 and a = 2(k + 1), b = k(k + 2), c = k 2 + 2k + 2.

NEW GENERAL FORMULAS FOR PYTHAGOREAN TRIPLES 3 Corollary 2.1. If γ = 1, 2 then prmtve Pythagorean trples are generated by γ = 1, β = 2n 2, a = 2n + 1 γ = 2, β = n 2, a = 4n b = 2n(n + 1) c = 2n 2 + 2n + 1 b = 4n 2 1 c = 4n 2 + 1 Proof. From Theorem2, f γ = 1, β = 2k 2 then a b, a c, b c and gcd(a, b, c) = 1. Thus prmtve Pythagorean trples can be found for k = n. If γ = 2, β = k 2 then we need to consder when k s even and when t s odd. If k s even let k = 2n then a = 2(2n+1), b = 4n(n+1), c = 2(2n 2 +2n+1) thus a b, a c, b c and gcd(a, b, c) = 2. Hence non-prmtve Pythagorean trples are found when k s even. If k s odd let k = 2n 1 then a = 4n, b = 4n 2 1, c = 4n 2 + 1 thus a b, a c, b c and gcd(a, b, c) = 1. Hence prmtve Pythagorean trples can be found when k s odd. From these results the corollary s proved. 3. General formulas for Pythagorean Trples Theorem 3. If (a, b, c) s a Pythagorean trple then there exsts α, β, γ Z, k, m, n N where α = b a, β = c a, γ = c b, beta = α + γ, (a γ) 2 = 2γβ, γβ = 2k 2, k = mn, m = 1, 2, 3,..., n = 1, 2, 3,... such that γ = m 2, β = 2n 2, a = m(m + 2n) b = 2n(n + m) c = m 2 + 2mn + 2n 2 γ = 2m 2, β = n 2, a = 2m(m + n) b = n(n + 2m) c = 2m 2 + 2mn + n 2 Proof. If n N, n = 1, 2, 3,... then n = 1(1), 1(2), 2(1), 3(1), 2(2), 5(1),2(3),..., 41(3), 31(4),... We see that n can be expressed as the product of two natural numbers. Therefore f k, m, n N and k = mn where m = 1, 2, 3,..., n = 1, 2, 3,... then snce γβ = 2k 2 we have γβ = 2m 2 n 2. At ths pont, we see that (γ, β) s {(m 2, 2n 2 ), (2m 2, n 2 )}. Thus by Theorem1 we get the general formulas. Observe that f m = 1 they become the formulas n Theorem 2. Corollary 3.1. If γ = m 2 and β = 2n 2 then prmtve Pythagorean trples are generated f m = 1 and m > 1 where m n, m s odd and gcd(m, n) = 1. Proof. If m = 1 they become the formulas n Theorem 2. If m > 1 and f q, t N, q = 1, 2, 3,..., t = 1, 2, 3,..., γ = m 2, m n, m = t then let m = 2t 1 and we have a = (2t 1)[(2t 1) + 2n], b = 2n[n + (2t 1)], c = (2t 1) 2 + 2n[n + (2t 1)], gcd(a, b, c) = 1. Now let m = 2t and we have a = 4(t)(t+n), b = 2(n)(2t+n), c = 2[2t 2 +2tn+n 2 ], gcd(a, b, c) = 2. If n = qm, then a = 2(1+q)m 2, b = q(2+q)m 2, c = (2+2q+q 2 )m 2. We see that gcd(a, b, c) = m 2 therefore consderng all of these, we conclude that prmtve Pythagorean trples are found f m = 1 and f m > 1, m n, m s odd and gcd(m, n) = 1.

4 EDUARDO CALOOY ROQUE Corollary 3.2. If γ = 2m 2 and β = n 2 then prmtve Pythagorean trples are generated f m = 1 and m > 1 where m n, n s odd and gcd(m, n) = 1. Proof. If m = 1 they become the formulas n Theorem 2. If m > 1 and f q, t N, q = 1, 2, 3,..., t = 1, 2, 3,..., γ = m 2, m n, m = t then let m = 2t 1 and we have a = 2(2t 1)[(2t 1)+n], b = n[n+2(2t 1)], c = 2(2t 1) 2 +n[n+2(2t 1)], gcd(a, b, c) = 1. Now let m = 2t and we have a = 2(2t)[2t+n], b = n[n+2(2t)], c = 2(2t) 2 +n[n+2(2t)], gcd(a, b, c) = 1. Snce both have gcd(a, b, c) = 1 we consder n. A quck mental calculaton wll tell us that gcd(a, b, c) = 2 f n s even but gcd(a, b, c) = 1 f n s odd. If q N, n = qm, q = 1, 2, 3,... then a = (2 + q)m 2, b = q(q + 2)m 2, c = (2 + 2q + q 2 )m 2. We see that gcd(a, b, c) = m 2 thus consderng all of these we conclude that prmtve Pythagorean trples are found f m = 1 and f m > 1, m n, n s odd and gcd(m, n) = 1. Corollary 3.3. If a < b then n > m 2 for γ = m 2, β = 2n 2 and n > m 2 for γ = 2m 2, β = n 2. Proof. If a < b then α > 0 and so β > γ. Thus for γ = m 2, β = 2n 2 we have m 2 > 2n 2 whch s n > m 2. And for γ = 2m 2, β = n 2 we have 2m 2 > n 2 whch s n > m 2. 4. Extra Theorem 4. If (a, b, c) s a Pythagorean trple and t N, t = 3, 4, 5,... then a t + b t c t. Proof. By Theorem 1 and the Bnomal Theorem we have Thus by Theorem 3 a t + b t c t (γ + 2k) t + (β + 2k) t [(γ + β) + 2k] t (γ) (t ) (2k) + (β) (t ) (2k) (γ + β) (t ) (2k) γ = m 2, β = 2n 2, (m 2 ) (t ) (2mn) + γ = 2m 2, β = n 2, (2m 2 ) (t ) (2mn) + (2n 2 ) (t ) (2mn) (n 2 ) (t ) (2mn) (m 2 + 2n 2 ) (t ) (2mn) (2m 2 + n 2 ) (t ) (2mn) where Z and k, m, n, t N, k = mn, t = 3, 4, 5,..., m = 1, 2, 3,..., n = 1, 2, 3,... Corollary 4.1. If (a, b, c) s a Pythagorean trple then a 3 + b 3 c 3.

NEW GENERAL FORMULAS FOR PYTHAGOREAN TRIPLES 5 Proof. From Theorem 4, f t = 3 then γ = m 2, β = 2n 2 a 3 = m 6 + 6m 5 n + 12m 4 n 2 + 8m 3 n 3 b 3 = 8n 6 + 24n 5 m + 24n 4 m 2 + 8n 3 m 3 c 3 = m 6 + 6m 5 n + 18m 4 n 2 + 32m 3 n 3 + 36m 2 n 4 + 24mn 5 + 8n 6 γ = 2m 2, β = n 2 a 3 = 8m 6 + 24m 5 n + 24m 4 n 2 + 8m 3 n 3 b 3 = n 6 + 6n 5 m + 12n 4 m 2 + 8n 3 m 3 c 3 = 8m 6 + 24m 5 n + 36m 4 n 2 + 32m 3 n 3 + 18m 2 n 4 + 6mn 5 + n 6 Thus we have for γ = m 2, β = 2n 2 a 3 + b 3 = m 6 + 6m 5 n + 12m 4 n 2 + 16m 3 n 3 + 24m 2 n 4 + 24mn 5 + 8n 6 c 3 = m 6 + 6m 5 n + 18m 4 n 2 + 32m 3 n 3 + 36m 2 n 4 + 24mn 5 + 8n 6 and for γ = 2m 2, β = n 2 a 3 + b 3 = 8m 6 + 24m 5 n + 24m 4 n 2 + 16m 3 n 3 + 12m 2 n 4 + 6mn 5 + n 6 c 3 = 8m 6 + 24m 5 n + 36m 4 n 2 + 32m 3 n 3 + 18m 2 n 4 + 6mn 5 + n 6 We see that a 3 + b 3 c 3 n both sets. The 3rd, 4th, and 5th terms dffer. Corollary 4.2. If (a, b, c) s a Pythagorean trple then a 4 + b 4 c 4. Proof. From Theorem 4, f t = 4 then γ = m 2, β = 2n 2 a 4 = m 8 + 8m 7 n + 32m 6 n 2 + 32m 5 n 3 + m 4 n 4 b 4 = 16n 8 + 64n 7 m + 128n 6 m 2 + 64n 5 m 3 + 16n 4 m 4 c 4 = m 8 + 8m 7 n + 32m 6 n 2 + 80m 5 n 3 + 136m 4 n 4 + 160m 3 n 5 + 128m 2 n 6 + 64mn 7 + 16n 8 γ = 2m 2, β = n 2 a 4 = 16m 8 + 64m 7 n + 128m 6 n 2 + 64m 5 n 3 + 16m 4 n 4 b 4 = n 8 + 8n 7 m + 32n 6 m 2 + 32n 5 m 3 + 16n 4 m 4 c 4 = 16m 8 + 64m 7 n + 128m 6 n 2 + 160m 5 n 3 + 136m 4 n 4 + 80m 3 n 5 + 32m 2 n 6 + 8mn 7 + n 8 Thus we have for γ = m 2, β = 2n 2 a 4 + b 4 = m 8 + 8m 7 n + 32m 6 n 2 + 32m 5 n 3 + 17m 4 n 4 + 64m 3 n 5 + 128m 2 n 6 + 64mn 7 + 16n 8 c 4 = m 8 + 8m 7 n + 32m 6 n 2 + 80m 5 n 3 + 136m 4 n 4 + 160m 3 n 5 + 128m 2 n 6 + 64mn 7 + 16n 8 and for γ = 2m 2, β = n 2 a 4 + b 4 = 16 m 8 + 64m 7 n + 128m 6 n 2 + 64m 5 n 3 + 32m 4 n 4 + 32m 3 n 5 + 32m 2 n 6 + 8mn 7 + n 8 c 4 = 16m 8 + 64m 7 n + 128m 6 n 2 + 160m 5 n 3 + 136m 4 n 4 + 80m 3 n 5 + 32m 2 n 6 + 8mn 7 + n 8 We also see that a 4 + b 4 c 4 n both sets. The 4th, 5th, and 6th terms dffer.

6 EDUARDO CALOOY ROQUE 5. Concluson We have found general formulas for Pythagorean trples and shown that they are vald. A glance between them and the formulas studed by the Greeks shows that the latter s a specal case. It s also evdent from these formulas that Pythagorean trples are nfnte and grouped nto two nfnte sets. As an extra, we found an applcaton for the formulas. It was shown that they could be used to prove that a t + b t c t for t 3 f (a, b, c) s a Pythagorean trple. Proofs for cases, t = 3, 4 were shown ndcatng that hgher values for t s also vald. The reason s that the terms around the mddle of the bnomal expansons wll always dffer for all t. A C++ scrpt that can be run n Clng and ROOT C/C++ nterpreters s attached here nstead of typesettng t verbatm.it s just a smple nteractve command lne nterface program. I also attached tables for γ = 1, 2, 2(2) 2, 3 2, 77 2, 2(77 2 ) wth n up to 10000 here: Wth hope that ths humble work be of beneft to fellowmen, we conclude wth these words: Proverbs 3:13 Happy s the man that fnds wsdom, and the man that gets understandng. Proverbs 9:10 The fear of the Lord s the begnnng of knowledge: and the knowledge of the Holy One s understandng.

NEW GENERAL FORMULAS FOR PYTHAGOREAN TRIPLES 7 6. Screenshots of C++ scrpt Fgure 1. S et 1 : γ = m 2, β = 2n 2

8 EDUARDO CALOOY ROQUE Fgure 2. S et 2 : γ = 2m 2, β = n 2

NEW GENERAL FORMULAS FOR PYTHAGOREAN TRIPLES 9 References [1] James Tattersall: Elementary Number Theory n Nne Chapters, (1999) [2] G. H. Hardy, E.M. Wrght: An Introducton to the Theory of Numbers, 4th ed., (1960) [3] M. Rchardson: College Algebra, 3rd ed., (1966) [4] Bjarne Stroustrup: C++ Programmng Language, 3rd ed. (1997) [5] Rene Brun, Fons Rademakers: ROOT User s Gude, (2007) [6] Vassl Vasslev: Clng The LLVM-based nterpreter, (2011) Phlppnes E-mal address: eddeboyroque@gmal.com

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