FLORENTIN SMARANDACHE A Method to Solve the Diophantine Equation ax 2 by 2 + c = 0

Transcription

1 FLORENTIN SMARANDACHE A Method to Solve the Diophatie Equatio ax 2 by 2 + c = 0 I Floreti Smaradache: Collected Papers, vol. I (secod editio). A Arbor (USA): IfoLearQuest, 2007.

2 A METHOD TO SOLVE THE DIOPHANTINE EQUATION ax 2 by 2 + c = 0 ABSTRACT We cosider the equatio (1) ax 2 by 2 + c = 0, with a,b N * ad c Z *. It is a geeralizatio of the Pell s equatio: x 2 Dy 2 = 1. Here, we show that: if the equatio has a iteger solutio ad a b is ot a perfect square, the (1) has a ifiitude of iteger solutios; i this case we fid a closed expressio for (x, ), the geeral positive iteger solutio, by a origial method. More, we geeralize it for ay Diophatie equatio of secod degree ad with two ukows. INTRODUCTION If ab = k 2 is a perfect square ( k N ) the equatio (1) has at most a fiite umber of iteger solutios, because (1) become: (2) (ax ky)(ax + ky) = ac If (a,b) does ot divide c, the Diophatie equatio does ot have solutios. METHOD TO SOLVE. Suppose that (1) has may iteger solutios. Let (x 0, y 0 ), (x 1, y 1 ) be the smallest positive iteger solutios for (1), with 0 x 0 < x 1. We costruct the recurret sequeces: x +1 = α x + β (3) +1 = γ x + δ makig coditio (3) verify (1). It results: aαβ = bγδ (4) aα 2 bγ 2 = a (5) aβ 2 bδ 2 = b (6) havig the ukows α, β, γ, δ. We pull out aα 2 ad aβ 2 from (5), respectively (6), ad replace them i (4) at the square; we obtai aδ 2 bγ 2 = a (7). We subtract (7) from (5) ad fid: α =±δ (8). Replacig (8) i (4) we obtai: β =± b a γ (9). Afterwards, replacig (8) i (5), ad (9) i (6) we fid the same equatio: aα 2 bγ 2 = a (10). Because we work with positive solutios oly, we take 16

3 x +1 = a 0 x + b a γ 0 +1 = γ 0 x + α 0 where (a 0,γ 0 ) is the smallest, positive iteger solutio of (10) such that a 0 γ 0 0. Let b α 0 a γ 0 M 2(Z). It is evidet that if (x', y') is a iteger solutio for (1) the γ 0 α 0 A x' y', x' A 1 y' is aother oe where A 1 is the iverse matrix of A, i.e. A 1 A = A A 1 = I (uit matrix). Hece, if (1) has a iteger solutio it has a ifiity. (Clearly A 1 M 2 (Z) ). The geeral positive iteger solutio of the equatio (1) is: GS 1 = ( ) ' ' ( x, y) x, y ( ) with x = A x 0 y 0, for all Z, where by covetio A 0 = I ad A k = A 1...A 1 of k times. I problems it is better to write GS GS 2 x ' ' ( ) ad x = A " " x 0 y 0 = A, N x 1 y 1 ( ) as:, N* ( ) is a geeral positive iteger We prove, by reductio at absurdum that GS 2 solutio for (1). Let (u,v) be a positive iteger particular solutio for (1). If x k 0 N :(u,v) = A k 0 0, or k x 1 N* :(u,v) = A k 1 1 the (u,v) ( GS 2). Cotrary to y 0 u i this, we calculate (u i+1,v i+1 ) = A 1 v i, for i = 0,1,2,... where u 0 = u, v 0 = v. Clearly u i+1 < u i for all i. After a certai rak x 0 < u i0 < x 1 it fids either 0 < u i0 < x 0, but that is absurd. It is clear that we ca put x ( GS 3 ) = x 0 A εy 0, N, where ε =±1. Now we shall trasform the geeral solutio GS 3 y 1 ( ) i a closed expressio. 17

4 Let λ be a real umber. Det(A λ I) = 0 ivolves the solutios λ 1,2 ad the v1 proper vectors V 1,2 (i.e., Av i = λ i v i, i { 1, 2} ). Note P = M 2( ) v2 The P 1 AP = λ λ 2, whece A = P λ 1 0 P 1, ad replacig it i GS 3 0 λ 2 ad doig the computatios we fid a closed expressio for ( GS 3 ). EXAMPLES 1. For the Diophatie equatio 2x 2 3y 2 = 5 we obtai x = ε, N ad λ 1,2 = 5 ± 2 6, v 1,2 = ( 6, ±2), whece a closed expressio for x ad : x = 4 + ε 6 ( ) + 4 ε 6 (5 2 6) 4 4 = 3ε ( ) + 3ε 2 6 for all N (5 2 6) For equatio x 2 3y 2 4 = 0 the geeral solutio i positive iteger is: x = (2 + 3) + (2 3) = 1 3 (2 + 3) + (2 3) that is (2,0), (4,2), (14,8), (52,30), EXERCICES FOR RADERS: Solve the Diophatie equatios: 3. x 2 12y = 0 x [Remark: = ε 4. x 2 6y 2 10 = 0 [Remark: x = x 2 12y 2 9 = 0 [Remark: x = x 2 3y 2 18 = 0 =?, N ] 4 ε =?, N ] 3 ε =?, N ] for all N, i ( ) 18

5 GENERALIZATIONS If f (x, y) = 0 is a Diophatie equatio of secod degree ad with two ukows, by liear trasformatio it becomes (12) ax 2 + by 2 + c = 0, with a,b,c Z. If ab 0 the equatio has at most a fiite umber of iteger solutios which ca be foud by attempts. It is easier to preset a example: 7. The Diophatie equatio (13) 9x 2 + 6xy 13y 2 6x 16y + 20 = 0 becomes (14) 2u 2 7v = 0, where (15) u = 3x + y 1 ad v = 2y + 1 We solve (14). Thus: u +1 = 15u + 28v (16), N with (u v +1 = 8u + 15v 0,v 0 ) = (3,3ε) First solutio: By iductio we prove that for all N we have that v is odd, ad u as well as v are multiple of 3. Clearly v 0 = 3ε, u 0. For + 1 we have: v +1 = 8u + 15v = eve + odd = odd, ad of course u +1,v +1 are multiples of 3 because u,v are multiple of 3 too. Hece, there exist x, i positive itegers for all N : x = (2u v + 3) / 6 (17) = (v 1) / 2 (from (15)). Now we ll fid the ( GS 3 ) for (14) as closed expressio, ad by meas of (17) it results the geeral iteger solutio of the equatio (13). Secod solutio: Aother expressio of the ( GS 3 ) for (13) will be obtaied if we trasform (15) as u = 3x + 1 ad v = for all N. Whece, usig (16) ad doig the computatio, we fid x (18) +1 = 11x + 11x N, with (x 0, y 0 ) = (1,1) or (2, 2) +1 = 12x (two ifiitude of iteger solutios) x 1 Let A = , the y = A 1 or

6 x 2 (19) y = A 2, always N. 1 1 From (18) we have always y 1 y... + y 0 1(mod 3), hece always x Z. Of course, (19) ad (17) are equivalet as geeral iteger solutio for (13). [The reader ca calculate A (by the same method liable to the start o this ote) ad fid a closed expressio for (19).]. More geerally: This method ca be geeralized for the Diophatie equatios: 2 (20) a i X i = b, with all a i,b Z. i=1 If always a i a j 0, 1 i < j, the equatio (20) has at most a fiite umber of iteger solutios. Now, we suppose i 0, j 0 1,..., { } for which a i0 a j0 < 0 (the equatio presets at least a variatio of sig). Aalogously, for N, we defie the recurret sequeces: (21) x (+1) () h = α ih x i, 1 h i=1 cosiderig (x 0 1,..., x 0 ) the smallest positive iteger solutio of (20). Replacig (21) i (20), it idetifies the coefficiets ad it looks for 2 ukows α ih, 1,i,h. (This calculatio is very itricate, but it ca be doe by meas of a computer.) The method goes o similarly, but the calculatios become more ad more itricate for example to calculate A, oe must use a computer. (The reader will be able to try this for the Diophatie equatio ax 2 + by 2 cz 2 + d = 0, with a,b,c N * ad d Z ) REFERENCES [1] M. Becze - Applicaţii ale uor şiruri de recureţă î teoria ecuaţiilor Diophatie - Gamma (Braşov), XXI-XXII, Aul VII, Nr. 4-5, 1985, pp [2] Z. I. Borevich, I.R. Shafarevich - Teoria umerelor - EDP, Bucharest, [3] A. Kestam - Cotributios to the Theory of the Diophatie Equatios Ax 2 By 2 = C. [4] G. H. Hardy ad E. M. Wright - Itroductio to the theory of umbers - Fifth editio, Claredo Press, Oxford, [5] N. Ivăşchescu - Rezolvarea ecuaţiilor î umere îtregi - This is his work for obtaiig the title of professor grade 2, (coordiator G. Vraciu), Uiv. Craiova, [6] E. Ladau - Elemetary Number Theory - Chelsea,

7 [7] Calvi T. Log - Elemetary Itroductio to Number Theory - D. C. Heath, Bosto, [8] L. J. Mordell - Diophatie equatios - Lodo, Academia Press, [9] C. Staley Ogibvy, Joh T. Aderso - Excursios i umber theory - Oxford Uiversity Press, New York, [10] W. Sierpiski - Oeuvres choisiers - Tome I, Warszawa, [11] F. Smaradache - Sur la resolutio d équatio du secod degré a dex icoues das Z i the book Gééralizatios et gééralités Ed. Nouvelle, Fes, Marocco; MR: 85, H: [Published i Gazeta Matematică, Serie 2, Vol. 1, Nr. 2, 1988, pp ; Traslated i Spaish by Fracisco Bellot Rasado, U metodo de resolucio de la ecuacio diofatica, Madrid. 21