Some Basic iophatie Equatios R.Maikada, epartmet of Mathematics, M.I.E.T. Egieerig College, Tiruchirappalli-7. Email: maimaths78@gmail.com bstract- - I this paper we preset a method for solvig the iophatie equatio, first we fid the polyomial solutio for the PELL S equatio by the method of cotiued fractios the preset the itegral solutio of the iophatie equatios. Theorems are discussed to demostrate the cotiue fractio method. Keywords- iophatie equatios, PELL S equatios ad Cotiue fractio method. I. INTROUCTION equatio which has two or more tha two ukows is called a idetermiate equatio. More geerally, a system of equatios is called idetermiate if the umber of equatios is less tha that of the ukows. iophatus, oe of the last lexadria Mathematicias of 3 rd cetury devoted himself to algebra. He proposed may idetermiate problems i this rithmetic. He was cotet with a sigle umerical ratioal solutio, although the problems usually have ifiitely may ratioal solutio ad ofte itegral solutios. Because he some times restricted his solutios to iteger, i his hoour, his ame is attached to the kid of idetermiate equatios for which the values of the variables are itegers ad study of such equatio is kow as IOPHNTINE NLYSIS. Oe ca easily uderstad that iophatie problems offer ulimited field for research by reaso of their variety. This dissertatio cosists of three chapters. I chapter, we preset of polyomial solutios for by the PELL s EQUTIONS employig the method of cotiued fractios. I chapter, we preset itegral solutios of the iophatie Equatios h h h h i) B C ( h, ) h h h h h ii) B C ( h, ) I chapter 3, we are cosider the famous Fermat s theorem ad preset a elegat way of provig the above theorem. II. POLYNOMIL SOLUTIONS FOR THE PELL S EQUTION INTROUCTION Let be a give square free atural umber. It is well kow that the Pell s equatios. () always has a ifiite umber of itegral solutios of () with positive ad B are obtaied by the formula r B r a b where r,,3,... ad a b r is the fudametal solutio of () a ad B b. The equatios. () may ot always have itegral solutios. Whe this equatio is solvable i iteger for some give, if be its fudametal solutios. Therefore we have the relatio a b L M. (3) where L M a b is the fudametal solutio of (). For a give value of, i order to determie the fudametal solutio of () or () (whe it exists). Oe may use the method of cotiued fractio expasio of. The peultimate coverget of this expasio provides the fudametal solutios. From computatioal poit of view, it would be advatageous to group value of ito several classes such that the values of i a specific class are associated with a similar patter of cotiued fractio expasios. It is foud that polyomial expressios for come i hady for this purpose. s a cosequece, for some polyomial solutios are give i this paper, for the Pell s equatios. FOR THE PELL S EQUTION Theorem Suppose t 5 where t is ay atural umber the a b is the fudametal solutio of. For real x, let the symbol x deote the greatest iteger x. We have ( t ) (t 3) ad so t. Now, 3 Iteratioal Joural of Iovative Kowledge Cocepts, 6(5) May, 08
expadig ito a cotiued fractio expasio, we obtai 4 t t 5 ( t ) (t ) (t ) (t ) (t ) t,, t,, t,, 4 t t t t t t t 3 t t 3 t t t t 6t 4 3 8t 9t 9 t 6t 4 t 5 (t ) t 5 (t ) t 5 (t ) 8t 4 t 4 t 5 (t ) 3 8t 9t 9. Theorem : Suppose 4 3 t 5 t 6t 4 m t mt t mt where m,,t are ay atural umbers with m0. The a b is the fudametal solutios of B where 3 a mt t, b t. For real x, let the symbol [x] deote the greatest iteger x. We have mt t mt t ad so [ ]. Now, expadig i to a cotiued fractio expasio. 4 3 mt t m t mt t mt ( mt t) mt ( mt t) t + 4 3 m t mt t mt mt t ( mt t) + 4 3 t m t mt t mt mt t mt... mt t, t,mt t mt t t 3 mt t t 3 4 3 mt t m t mt t mt ( t) Theorem 3: Suppose m t 4 mt 3 t mt where m,, t are ay atural umbers with m 0. The a b is the fudametal solutio of 3 B where a mt t, bt. Proof : Cosider [] mt t ad do the proof by the cotiued fractio doig the above theorem, we get the expasio of mt t, t,mt t mt t t 3 3 mt t m t 4 mt 3 t mt t FOR THE PELL S EQUTION Theorem 4: If 5 where t is ay atural umber the L M is the fudametal solutio of 3 where L 6t 6t, m t t 4 Iteratioal Joural of Iovative Kowledge Cocepts, 6(5) May, 08
For real x, let the symbol [x] deote the greatest iteger x. We have (t+) < [ ] t+ Now, expadig freedom fractio expasio. We obtai <(t+).. ad so t t t [ ] 4 4 5 t t (t ) t... [t+,t,,,t,+] t t t 3 6t 6t t t Thus we get the solutio ito a cotiued 4 4 5 t t t 5 t 3 t t t t t t t 4 6 6 4 4 5 Theorem 5: 4 3 If 6t 8t 9t 6t where t is ay atural umber the L M is the fudametal solutio of where 4 3 L 6t 8t 3t, m Cosider t ad do the proof by the expasio of i a cotiued fractio we obtai t, t, t,8t t 4 3 6t 8t 3t 4 3 4 3 t t t t t t t t t 6 4 8 3 6 8 9 6 4 Theorem 6: If, where t is ay atural umber, the L M is the fudametal solutio of where 4 3 L t M t t. Cosider t ad do the proof by the expasio of i a cotiued fractio are obtai t, t,,, t, 4 3 t t t t 4 3 4 t t t t t t t t 4 4 4 4 THE GENERL PELL S EQUTIONS Theorem 7: Let u, v be iteger such that u v N, where N is ay give o zero iteger. The Kv u,, B V K V ku satisfy B N, Where K is ay atural umber Give Kv u B () () V K v Ku where K is ay atural () umbers choose for K ay value satisfyig B N from the help of above theorem, oe may geerate the solutio of the geeral Pell s equatio. u v N where N is a o zero iteger Corollary : Let K, t be atural umber the Kt, B t, K t k satisfy B THE IOPHNTINE EQUTION 4 The solutio of the iophatie equatios 4 may be geerated from the solutio of the Pell s equatio u v. Theorem 8 : Let u,v be iteger such that k u V B V, k K u v. The u 4 satisfy 5 Iteratioal Joural of Iovative Kowledge Cocepts, 6(5) May, 08
Give: K u v BV u 4 k Therefore it satisfies k k u u v 4V u v k u v k u v u k v k Whe k, the give coditios u vk reduces to PELL S Equatio u v with ad hece if possesses ad ifiite umber of solutios for both sigs. Thus there are a ifiite umber of values of for which be equatio 4 has o- trivial iteger solutios, as provided by the give statemet. However, whe k 3, by Thue s Theorem (Mordell), u vk has atmost a fiite umber of itegeral solutios. For example i the case of k4, Ljuggre proved that the oly the iteger solutio of the iophatie equatio. 4 u v re (u,v) (,) ad (39,3). Usig the better solutio of this equatio i the give theorem, we obtai the relatio, 68638-575(3) 8 - Thus providig a solutio of the iophatie equatio. with 575 8 lso cosiderig fially, kowig the solutios, B B, the solutios for the 0 0 equatio are give by, B. 0 0 SOLUTION OF IOPHNTINE EQUTION method of costructig special solutio of iophatie Equatios: h h h h B c h, i) h h h h h ii) B c h, has bee described. We kow that every prime p (mod6) (or) every prime of the form 6k+ ca be expressed i the form p pq q () where p,q are +ve iteger lso eve iteger N which is the product of primes of the form 6K+ ca be represeted i the form () repeated applicatios of the formula. a ab b c cd d x xy y () Where xac+bc+bd, yac+ad+bd we therefore observe that the product of primes of the form 6K+ is agai a prime of the form 6K+. For example: 7 (mod 6), 9 ( mod 6) 7x9 33 (mod 6) If N ca be expressed i the form () the, above beig atural umber > ca also be expressed i the same form usig the idetity () Now we cosider the iophatie Equatios h h h h N x y z h, where (3) N p pq q Settig x -pq ad substitutig h, we have i tur i equatio (3), we obtai respectively,. N y z N x p pq q pq p q (4a) y z N x N x N x p q p q (4b) Cosider the idetity ( ) y z y z y z Usig equatio (4a), (4b) we have 4 y z p q p q p q ( ) p q p q p q p q p q pq p q p q ( y z) p q (5) Takig +ve square root y z p q (5a) From (4a) & (5a) we obtai y p pq, z q pq Substitutig the values of x,y,z i (3) we obtai the idetities p pq q ( pq) p pq q pq p pq q ( pq) p pq q pq (6) Note that we obtai the same equatio (6) whe we substitute the ve square root i equatio (5) 6 Iteratioal Joural of Iovative Kowledge Cocepts, 6(5) May, 08
If N C p pq q where C has oly prime factors of the form 6K+5 or, we multiply the system (3) by C ad obtai the same idetities (6) lso the N C p pq q, C P p q q where C,, iteger N C P qp q p q are Thus there exists itegers x,y,z such that p qp q x y z ad p pq q x y z N C p p q q c ( x y z c x c y c z & ( N C p p q q c x y z c x c y c z idetities (6) ad so for (ii), (iii), (iv) The above aalysis leads to the followig result stated as Theorem : If the iteger >0 ad N a atural umber havig at least oe prime factor of the form 6k+, the diophatie equatios. N x x x 3 N x x x3 have always a solutio. Result : p pq q remais ualtered if we replace p by p+q ad q by q or if we replace p by p+q ad q by p (i.e) p pq q p q p q q q p q p q p p For each of the above substitutios for p ad q the 3 umbers pq, p pq, q pq are oly permutatios leadig to the same idetities (6). Result : If we iterchage either a, b or c, d (but ot both simultaeously) i the form () we get a ew set of values of & B (which is the same for either iterchage) through the product of LHS of () does ot chage. Example : N 47 (3) (9) ( +()(3)+3 ) ( +()(3)+3 ) 7 - (7)(4)+4 3 +3(4)+ 4 (i) (3 +3()+ )( +(3)+3 ) -(8)+8 7 +7()+ (ii) ad we have 47 (-4)+5+38 (-77) +6+98 47 (-4) +5 +38 (-77) +6 +98 47 (47) (47) (3 +3(4)+4 ) (7 +7()+ ) (3 +3(4)+4 )(7 +7()+ ) (4 +4(3)+3 ) (7 +7()+ ) (87 +87(93)+93 ) (i) 7 +7(03)+03 (ii) 65 +65(08)+08 (iii) 33 +33(5)+5 (iv) ad with (i) we have 47 (-739) +5360+6040+ 47 (-739) +5360 +6040 (from gai takig (ii) ad (i) we have 47 3 47 (47) (87 +87(93)+93 )(7 +7()+ ) (983-983(4389)+4389 ) 406 +406 (983)+983 (i3) 86 +86(373)+373 (ii3) ad with (i3) we have 47 3 (-494098)+670934+309387 47 6 (-494098) +670934 +309387 etc., The idetities (6) ca be writte as (p +pq+q )+(p +pq+q ) p +q +(p+q) (7) (p +pq+q ) +(p +pq+q ) p 4 +q 4 +(p+q) 4 (8) Hece we have Theorem : If the iteger >0 & N a atural umber havig at least oe set of U, U, U 3 where U +U U 3 which satisfy the iophatie equatios. N N U U U3 N N U U U 4 4 4 3 Example : With (i) & (ii) of above ad previous we have 47+47 3 +4 +7 7 + +8 (47+47) 3 4 +4 4 +7 4 7 4 + 4 +8 4 Now we cosider the solutios of iophatie equatios for the value of h3 i the expasios h h h h N x y z where N p pq q Proceedig as above we have the values x-pq, y+z(p+q),y +z (p +q )(p+q) cosider y 3 +z 3 (y+z)[y -yz+z ] (p+q) [(p +q )(p+q) -yz] (9) y 3 +z 3 N 3 -x 3 from our assumptio (N-x) (N +Nx+x ) (p+q) [p +pq+q )(p +q )+p q ] (0) 7 Iteratioal Joural of Iovative Kowledge Cocepts, 6(5) May, 08
From equatio (9) & (0) yz pq (p +q )-p q Now cosider the idetity (y-z) (y+z) -4yz (p+q) 4-4pq(p +q )+4p q p 4 +4p 3 q+6p q +4pq 3 +q 4-4p 3 q- 4pq 3 +4p q p 4 +0p q +q 4 4 4 (y-z) p 0p q q lso y+z (p+q) Solvig the above two equatios we get the irratioal solutios of y & z. PROOF OF FERMT S LST THEOREM Fermat s last theorem is based uder the coditio x y z which is impossible for atural umbers where m is greater tha. (i.e) From the coditios x y z where x, y, z, m N ad m>. But i the followig three defiite relatios : (i) x y z (Pythagorea triplet) (ii) x y z (iii) x y z (Iequality) () Case () x y z, x z, y z () m x Multiply equatios () o both sides by m m m x x y x x z () Sice from the assumptio x<y<z i () m m m m y x z x We get lso m m m m y y x z z x y<z) (3) (sice ddig () & (3) we get the result x y z x y z B) It may be proved also, multiplyig () by xyz m x y z x xyz y xyz z xyz m m m m m m x yz y xz z xz Sice from the assumptio x y z, xz zx yz Now x y z is possible, xy zx yz m m m But it is impossible to our assumptios x y z. So x y z (C) Case : (4) x y z where Multiply equatios (4) o both sides by m m m x x y x x z (5) Sice from the assumptio x y z x y z We get m m m y x z x m lso m m m m y y x z z x y<z) (6) m x (sice ddig (5) & (6) we get x y z x y z () It also proved as from iequality x<y<z for our assumptio x y z x y z z x y z x y (sice m>) x y z (E) case : 3 x y z where (i) x>z & y>z (ii) z>x & y>z (iii)x>z & z>y The above iequality is the back boe for this problem. We have to prove that for our above assumptio i aother way. F (i.e) cosider x y z x<z & y<z Let it be assumed to be such that x y z alters to x y z for m3 If x, y, z are of α phythogoria triplet the x y z Multiplyig the above result by a o both sides ax ay az It may be proved as i a differet maer V (cuboid ) + V (Cuboid ) V 3(cuboid 3) of Square base x y z Height a a a Now x y z (i.e) V (cube ) + V (Cube ) V 3(cube 3) of Square base x y z Height x y z s cube may be steaped as cuboid we have V (cube ) V (cuboid ) ad so o. 3 3 3 x ax, y ay ; z az Certaily x, y, z d ad a x, y, z 3 3 3 x / a x, y / a y, z / a z such that x y z ad 8 Iteratioal Joural of Iovative Kowledge Cocepts, 6(5) May, 08
3 3 3 x / x y / y z / z But this impossible because x / x y / y z / z is impossible I this way our suppositio is false ad x y z as well as x y z (G) It may be proved as uder too x<y<z y 3 z x x y z For m3; (i) (ii) 3 y z x x y z I geeral y 3 z x x y z (i) (ii) 3 y z x x y z x y z CONCLUSION So for we have see how cotiued fractios solves the iophatie Equatios i the pplied Mathematics. Thus pplied Mathematics without Pure Mathematics has o root ad Pure Mathematics without pplied Mathematics leaves o fruit. REFERENCES. W. Ljuggre, avh. Norske, vid. kad.oslo, o.5 (94). L.j.mordell, diophatie equatios, academic press, lodo ad ewyork, 969, mr40, 600. 3. T.agell, itroductio to umber theory, wiley, ewyork, 95, mr 3, 07. 4. B.r.paday, a proof of fermat s last theorem, tme, volume xxix, o.4, december 995. 5. J.choubey, method of costructig special solutios of diophatie equatios, tme, volume, xxviii, o. march 994. 6..m.s. Ramasamy, polyomial, solutios of the pell s equatio, ijpam, volume 5, o.6, jue 994. 9 Iteratioal Joural of Iovative Kowledge Cocepts, 6(5) May, 08