Pythagorean triples. Leen Noordzij.

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1 Pythagorean trple. Leen Noordz Content A Roadmap for generatng Pythagorean Trple.... Pythagorean Trple Dcuon Concluon A Roadmap for generatng Pythagorean Trple. To fnd the oluton, expreed nto prme number, of a + b = c, (.) where a, b en c are potve nteger number, {a, b, c N a, b, c > 0}, a tard procedure. Intructve readng on Pythagorean Trple can be found, e.g., n a, b en c conttute a o-called Pythagorean Trple. In general a even b c are odd. The approach we chooe baed on The Fundamental Theorem of Arthmetc. The Theorem: any nteger greater than ether a prme number, or can be wrtten a a unque product of prme number. We wrte a a a product of prme number, : a l 3 n 5 n 7 n 3.., (.) wth n, n,,., 0 en poteve gehele getallen en l eveneen een potef geheel getal. Or a = l n =. Wth { N 0} {l N l > 0}. belong to the ubet of odd prme number P: { P odd} P N. Furthermore: + >. b en c can be expreed n a, wth (.), : (c + b)(c b) = a. (.3) c + b c b are even potve nteger. So, {c ± b N c ± b > 0 even}.

2 Eucld proofed the extence of anfnte number of Pythagorean trple. c + b c b are nteger factor of a. For example: wth a = 4, c + b = 8 c b =. Wth help of Eq. (.3) we have c + b < a, nce c b >. Alo c b < a, nce c + b >. Wth c + b c b to be even potve nteger, c + b c b are nteger factor of a ndeed. So c + b = a a, c b = c b. c+b n For convenence we wrte a = l = = l P, (.4) where P are product of power of odd prme number. P beng co-prme: P = {, P For = 0 : = P =. n = }. For 0 : a poble combnaton of {, P } : = n = P =, P = n = P =. The lowet poble trple obtaned for = 0 l = : (4, 3, 5). For any combnaton of b c, en P can be found from dfferent combnaton of product of power of prme number gven Eq. (.). In general, wth (.4), we have: (c + b)(c b) = a = l P. (.5) Snce b c are co-prme note the above repreentaton of a n (.4) not ut for convenence: c + b = l k, c b = k P. For c b we have: (.6) (.7) c = l k + k P, (.8) b = l k k P. (.9) {k N 0 < k < l}. Keep n mnd: c > b. To make c b co-prme ndeed, we need to do omethng addtonal, unle: k = 0 or l k = 0, (.0) we do not obtan a prmtve Trple. Furthermore, thee two contrant lead to the followng condton: l >. Snce for l =, the aforementoned contran, the rght-h de of Eq. (.8) (.9) repreent the um of two odd nteger become conequently even. So a, n the Trple you are lookng for, at leat anteger that can be dvded by 4. Agan: for l =, k = we fnd the Trple (4, 3,5). We alo need the contrant: b > 0. Wth (.9) we obtan: l k > k P > k P l. (.)

3 To ummarze: a = l n = l P, k P p P p n l l l p p n l l no Note: the mnmum value of l =. Pythagorean Trple. k Trple + c l ye/no b ye l ± P l l? l ± P l? l ± P ye l ± l l? ± l P l? ± l P l no Let u tart wth l =, = 0 wth (.0) k = or k = 3. - k =. We have a = 4, wth (.4) (.) P = > k l =. Wth (.8) (.9): c = 4 + = 5, b = 4 = 3. We obtan the trple we are famlar wth: (4, 3, 5). - k = 3. There no trple. The trple not allowed due the contrant (.). 3

4 Now for the next trple(): l =, =, =, = 3 wth (.0) k = or k = 3. - k =. We have the mallet prme: 3. = 3 P =. Then a = 4 3 =, wth (.4) (.) from (.8) (.9) we fnd: c = = 37, b = 4 9 = 35. The trple : (, 35, 37). - k = 3. wth (.4) (.) P = 3 > k l =. P = 3 > k l =. So we expect another trple. Wth (.8) (.9): c = = 3, b = 9 4 = 5. The trple (, 5, 3). So we have two trple for a = 4 3 =. However n the algorthm you need to fnd out about = P = 3. The contrant (.) forbd another trple. Let u try another one. Now for the next trple(): l =, =, =, = 5 wth (.0) k = or k = 3. - k =. We ue the econd mallet prme: 5. = 5 P =. Then a = 4 5 = 0, wth (.4) (.) from (.8) (.9) we fnd: c = = 0, b = 4 5 = k = 3. wth (.4) (.) So we expect another trple. Wth (.8) (.9): c = = 9, b = 5 4 =. The trple (0,, 9). P = 5 > k l =. P = 5 > k l =. 4

5 We have two trple for a = 0. For a partcular value of a there are more than non-trval or prmtve Pythagorean Trple. An example: a = 60(=. 3. 5) produce four Pythagorean Trple : (60., 6), (60, 9, 09), (60,, 9) (60, 899, 90). So, wth a = l n =, =, =, p = 3, p = 5 l = : k P /P k / l trple / Ye: 60, 899, /5 / Ye: 60,, /3 / Ye: 60, 09, /5 / no Ye: 60,, /5 no /3 no /5 no Four trple Note: gven a = 80(=. 3. 5), o n = you wll fnd agan four trple. Dcuon Concluon. In the ecton on Road Map we chooe: c + b = l k, c b = k P. 5 (.6) (.7) c = l k + k P, (.8) b = l k k P. (.9) Or more conce: ( (l k ) k P ) = (c b ). On the other h we could have choen: c + b = r, (3.) c b = l r P. Gvng: (3.) c = r + l r P, (3.3) b = r l r P, (3.4) {r N 0 < r < l}. Or more conce: ( ) ( r l r P ) = (c b ).

6 In order to fnd trple we have the condton: r = 0 or l r = 0. (3.5) In addton the followng contrant apple: b > 0. Wth (3.4) we obtan: r > l r P > l P r. (3.6) We expect the dtrbuton of the prme number gven (3.) (3.) to delver the ame reult for the Trple. Well, comparng the condton (3.5) wth Eq. (.8) (.9) we ee: (r ) to be equvalent wth (l k ), (l r ) to be equvalent wth (k ). Plug r = 0 nto (3.6) we fnd P > l (= (l ) ). (3.7) Now, plug (l k ) = 0 nto l k > k P we have: P > l l (= l ) ), > k P equvalent to (3.7). Plug l r = 0 nto (3.6) we fnd P > l l, (.) l (= /(l ) ). (3.8) Now, plug (k = 0) nto l k > k P > P l, (.) equvalent to (3.8). Hence, we fnd wth the dtrbuton of the prme number gven (3.) (3.) the ame Trple ndeed. For creatng the Pythagorean Trple you need an effcent algorthm for producng the prme number: Generatng Prme, Seve of Atkn, Queton: how many trple can we obtan for a gven value of a? A fnal remark. You wll notce not to fnd, e.g., a combnatouch a: (9,, 5). Th a non-prmtve trple. Well, et l =, = P, a = you wll fnd th non-prmtve oluton. In th cae by = P ( ) a non-prmtve oluton ha been forced. A mpler approach by multplyng the prmtve Trple (3, 4, 5) by any potve nteger you lke. The Trple (4, 3, 5) found wth = 0 n (.4) gvng P equal n (.8) (.9). 6

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Foundations of Arithmetic

Foundations of Arithmetic Foundatons of Arthmetc Notaton We shall denote the sum and product of numbers n the usual notaton as a 2 + a 2 + a 3 + + a = a, a 1 a 2 a 3 a = a The notaton a b means a dvdes b,.e. ac = b where c s an