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1 Pythagorean Triple Updated wwwleennoordzijme Content A Roadmap for generating Pythagorean Triple Pythagorean Triple 3 Dicuion Concluion 5 A Roadmap for generating Pythagorean Triple To find the olution, repreented by prime number, of a + b = c, () where a, b en c are poitive integer number, {a, b, c N a, b, c > 0}, i a tard procedure Intructive reading on Pythagorean Triple can be found, eg, in wwwenwikipediaorg a, b en c contitute a o-called Pythagorean Triple In general a i even b c are odd The approach we chooe i baed on The Fundamental Theorem of Arithmetic The Theorem: any integer greater than i either a prime number, or can be written a a unique product of prime number We write a a a product of prime number, : a l 3 n 5 n 7 n 3, () with n, n,,, 0 So a = l n = With { N 0} {l N l > 0} belong to the ubet of odd prime number P: { P odd} P N Furthermore: + > b en c can be expreed in a, with (), : (c + b)(c b) = a (3) c + b c b, {c ± b N c ± b > 0 even} Euclid proofed the exitence of an infinite number of Pythagorean triple

2 c + b c b are integer factor of a For example: with a = 4, c + b = 8 c b = With Eq (3) we have c + b < a, ince c b > Alo c b < a, ince c + b > With c + b c b to be even poitive integer, c + b c b are integer factor of a indeed So c + b = a a, c b = c b c+b n For convenience we write a = l = = l P j, (4) where P j are product of power of odd prime number P j being co-prime: P j = {, P j For = 0 : = P j = n = } For 0 : a poible combination of {, P j } i: = n = P j =, P j = n = P = Furthermore: < P j or P j < The lowet poible triple i obtained for = 0 l = : (4, 3, 5) For any combination of b c, en P j can be found from different combination of product of power of prime number given in Eq () In general, with (4), we have: (c + b)(c b) = a = l P j (5) Since b c are co-prime note the above repreentation of a in (4) i not jut for convenience With (5), c + b c b can be repreented repectively: c + b = l k P i, (6) c b = k P j For c b we have: (7) c = l k + k P j, (8) b = l k k P j (9) {k N 0 < k < l} Keep in mind: c > b c, b odd Conequently c b > (c + b) < l P j To make c b co-prime indeed, we need to do omething additional, unle: k = 0 or l k = 0, (0) we do not obtain a primitive Triple With k =, l k ha to be larger than 0, or l > 0 l >, with l k = 0 k = l k = l, hence l > So, thee two contraint lead to the following condition: l > Since for l =, the aforementioned contraint, the right-h ide of Eq (8) (9) repreent the um of

3 two odd integer become conequently even a, in the Triple you are looking for, i at leat an integer that can be divided by 4 Again: for l =, k = we find the Triple (4, 3,5) We alo need the contraint: b > 0 With (9) we obtain: l k > k P j > k P j l () To ummarize: a = l n = l P j, Table Roadmap: k P j p i n i i j i p j n j j j i P j p i n i i p j n j j n l l p i n i i j l i p j n j j j i p i n i i p j n j j k Triple + c l b ye l P i ± l? l P i ± P j l? l ± P j l ye l ± l l? P i ± l l? ± l P j l no l l no Note: the minimum value of l = Pythagorean Triple Let u tart with l =, = 0 with (0) k = or k = 3 - k = We have a = 4, 3

4 with (4) () P j = > k l = With (8) (9): c = 4 + = 5, b = 4 = 3 We obtain the triple we are familiar with: (4, 3, 5) - k = 3 There i no triple The triple i not allowed due the contraint () Now for the next triple(): l =, =, =, = 3 with (0) k = or k = 3 - k = We have the mallet prime: 3 = 3 P j = Then a = 4 3 =, with (4) () from (8) (9) we find: c = = 37, b = 4 9 = 35 The triple i: (, 35, 37) - k = 3 with (4) () P j = 3 > k l = P j = 3 > k l = So we expect another triple With (8) (9): c = = 3, b = 9 4 = 5 The triple i (, 5, 3) So we have two triple for a = 4 3 = However in the algorithm you need to find out about = P j = 3 The contraint () forbid another triple Let u try another one Now for the next triple(): l =, =, =, = 5 with (0) k = or k = 3 - k = We ue the econd mallet prime: 5 = 5 P j = Then a = 4 5 = 0, with (4) () from (8) (9) we find: c = = 0, P j = 5 > k l = 4

5 b = 4 5 = 99 - k = 3 with (4) () P j = 5 > k l = So we expect another triple With (8) (9): c = = 9, b = 5 4 = The triple i (0,, 9) We have two triple for a = 0 For a particular value of a there are more than non-trivial or primitive Pythagorean Triple An example: a = 60(= 3 5) produce four Pythagorean Triple : (60, 6), (60, 9, 09), (60,, 9) (60, 899, 90) So, with a = l n =, =, =, p = 3, p = 5 l = : k P j /P j k / l triple / Ye: 60, 899, /5 / Ye: 60,, /3 / Ye: 60, 09, /5 / no Ye: 60,, /5 no /3 no /5 no Four triple Note: - given a = 80(= 3 5), o n = you will find again four triple - given a = 3 57, you will find eight triple - given a = 3 57, you will find ixteen triple Dicuion Concluion In the ection on Road Map we chooe: c + b = l k, c b = k P j 5 (6) (7) c = l k + k P j, (8) b = l k k P j (9) Or more concie: ( (l k ) k P ) = (c j b )

6 On the other h we could have choen: c + b = r, (3) c b = l r P j Giving: (3) c = r + l r P j, (33) b = r l r P j, (34) {r N 0 < r < l} Or more concie: ( ) ( r l r P ) = (c j b ) In order to find triple we have the condition: r = 0 or l r = 0 (35) In addition the following contraint applie: b > 0 With (34) we obtain: r > l r P j > l P j r (36) We expect the ditribution of the prime number given in (3) (3) to produce the ame reult for the Triple Well, comparing the condition (35) with Eq (8) (9) we ee: (r ) to be equivalent with (l k ), (l r ) to be equivalent with (k ) Plug r = 0 into (36) we find P j > l (= (l ) ) (37) Now, plug (l k ) = 0 into l k > k P j we have: P j > l l (= l ) ), > k P j equivalent to (37) Plug l r = 0 into (36) we find P j > l l, () l (= /(l ) ) (38) Now, plug (k = 0) into l k > k P j then P j > l > k P j l, () equivalent to (38) Hence, we find with the ditribution of the prime number given in (3) (3) the ame 6

7 Triple indeed For creating the Pythagorean Triple you need an efficient algorithm for producing the prime number: Generating Prime, Sieve of Atkin, wwwenmwikipediaorg Quetion: how many triple can we obtain for a given value of a? Well, for a = l n =, for a particular value of l one triple i found See Table at the end of ection A road map for generating Pythagorean triple Thi i equivalent to the tatement: for = 0, one triple can be found Now we chooe a = l n p t t, ie one odd prime number nt We find at leat one n triple Since for = p t t Pj = : > P j l I there another triple? We have two poibilitie: k =, then = P j = p t n t, k = l, then = p t n t Pj = We analye k = : Plug into P j > l the value for P j : l > + n tln (p t ) ln Next k = l : l < + n tln (p t ) ln We conclude to find another triple for l > + n tln (p t ) ln or l < + n tln (p t ) ln Hence: for a = l p t n t, ie one odd prime number nt, we obtain two Pythagorean triple Conjecture: for a given a the number of triple equal, where i the number of odd-coprime-triple of which a i compoed Note: we ue the wording co-prime, ince for example 5 i a eparate number in the triple of odd prime number erie A final remark You will notice not to find, eg, a combinatiouch a: (9,, 5) Thi i a non-primitive triple Well, et l =, = P j, a = you will find thi non-primitive olution In thi cae by = P j ( ) a non-primitive olution ha been forced A impler approach i by multiplying the primitive Triple (3, 4, 5) by any poitive integer you like The Triple (4, 3, 5) i found with = 0 in (4) giving P j equal in (8) (9) 7

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