International Journal of Science and Research (IJSR) ISSN (Online): 2319-7064 Index Copernicus Value (2013): 6.14 | Impact Factor (2014): 5.611 On The Homogeneous Biquadratic Equation With 5 Unknowns: x4-y4=65 (z2-w2) R2

Dr. P. Jayakumar1, R. Venkatraman2

1Professor of Mathematics, Periyar Maniammai University, Vallam, Thanajvur-613 403,Tamil Nadu, India

2Ph.D. Scholar, Assistant Professor of Mathematics, SRM University Vadapalani Campus, Chennai -600026. Tamil Nadu, India

Abstract: The Homogenous biquadratic equation with five unknowns given by x4-y4=65 (z2-w2) R2 is considered and analyzed for finding its non zero distinct integral solutions. Introducing the linear transformations x = u + v, y = u – v, z =2uv +1, w =2uv-1 and employing the method of factorization different patterns of non zero distinct solutions of the equation under the above equation are obtained. A few interesting relations between the integral solutions and the special numbers namely Polygonal numbers, Pyramidal numbers, Centered Polygonal numbers, Centered Pyramidal numbers, Thabit-ibn-Kurrah number, , Carol number, woodall number, kynea number, ,stellaoctangul number,, Mersenne number are exhibited.

Keywords: Homogeneous equation, Integral solutions, Polygonal numbers, Pyramidal and special number.

2010 Mathematics Subject Classification: 11D09

Notations used: Tm,n - of rank n with size m Pm n - of rank n with size m gn - Gnomonic number of rank n

Prn - of rank n Ct16,n- Centered hexadecagonal pyramidal number of rank n OHn - Octahedral number of rank n SOn - Stella octangular number of rank n \kyn - kynea number carln -carol number

1. Introduction 2.1 Pattern: I

The theory of Diophantine equations offers a rich variety of Assume 65 = (8 + i) (8 – i) (4) 2 2 fascinating problems. In particular biquadratic Diophantine and R= a + b = (a + i b) (a – i b) (5) equations, homogeneous and non-homogeneous have Using (4) and (5) in (3) and employing the method of aroused the interest of numerous mathematicians since factorization we get. 2 2 antiquity [1-12]. In this context one may refer [4-10] for (u + i v) (u – iv) = (8 + i) (8 – i) (a + i b) (a – i b) various problems on the biquadratic Diophantine equations. However, often we come across non-homogeneous On equating the positive and negative factors, we have, 2 biquadratic equations and as such one may require its (u + i v) = (8 + i) (a + i b) 2 integral solution in its most general form. This paper (u + i v) = (8 – i) (a – i b) concern with the homogeneous biquadratic equation with five unknowns x4–y4=65(z2–w2)R2 for determining its On equating real and imaginary parts, we get 2 2 infinitely many non-zero integral solutions. Also a few u = u (a, b) = 8a – 8b – 2ab 2 2 interesting properties among the solutions are presented. v = v (a, b) = a – b + 16ab

2. Method of Analysis On substituting u and v in (2) we get the values of x, y, z and w. The non-zero distinct integrals values of x, y, z, w and R The biquadratic equation with five unknowns to be solved satisfying (1) are given by x = x (a, b) = 9a2 – 9b2 + 14ab for its non-zero distinct integral solution is 2 2 44 222 y = y (a, b) = 7a – 7b – 18ab  Rwzyx (1) 4 4 2 2 3 3 65 )( z = z (a, b) = 2(8a + 8b – 48a b + 126a b – 126ab ) +1 w = w (a, b) = 2(8a4 +8b4 – 48a2 b2 + 126a3b –126ab3) –1 Consider the transformations R = R (a, b) = a2 + b2 x = u + v, y = u – v, z = 2uv + 1, w = 2uv – 1 (2) Properties:- On substituting (2) in (1), we get 1. x (2, a) + y (2, a) + Ct16,a + 8t4,a – W4 = 0 u2 + v2 = 65R2 (3) 2. R (a+1, a+1) – 2t4,A – G2a 0 (mod3) Volume 5 Issue 3, March 2016 www.ijsr.net Paper ID: NOV162379 Licensed Under Creative Commons Attribution CC BY 1863 International Journal of Science and Research (IJSR) ISSN (Online): 2319-7064 Index Copernicus Value (2013): 6.14 | Impact Factor (2014): 5.611

3. z (1,b) – W(1,b) – OH2 On equating the rational and irrational parts, we get ( + 8) ( a + b)2 = ( R + v) 4. x [a(a+1),1] – 9y[a(a+1),1] – 260Pa = 0 2 2 2 5. 7x [a(2a -1),1]-9y[a(2a -1),1] – 260Sa = 0 ( 8) ( a b) = ( R - v) 5   PR 3  03,3.6 yx  )2(mod0)2,2()2,2(.7 On equating the real and imaginary parts, we get R = R (a, b) = 65a2 + b2 + 16ab 2  aayaax 2 OH  2 2 12(,1[7.8 )] 12(,1[9 )] a 0864 v = v (a, b) = 520a + 8b + 130ab

Twz 23 4,  )2(mod0)1,1()1,1(.9 2 Substituting the values of u and v in (2), the non – zero 10 1(,1[7. )] aayaax 1(,1[9 )] P  .0)(260 a distinct integral values of x, y, z, R and w satisfying (1) are given by 2.2 Pattern: II x = x (a, b) = 585a2 + 7b2 + 130ab y = y (a, b) = –455a2 – 9b2 – 130ab Also 65 can be written in equation (3) as z = z (a, b) =2(33800a4 – 8b4 - 130ab3 + 8450a3b) + 1 65 = (1 +8i) (1 – 8i) (6) w = w (a, b) = 2(33800a4 – 8b4 - 130ab3+ 8450a3b) -1 Using (5) and (6) in equation (3) it is written in factorizable R = R (a, b) = 65a2 + b2 + 16ab form as 2 2 (u + iv) (u – iv) = (1 +8i) (1 – 8i) (a + ib) (a – ib) Properties:

On equating the positive and negative factors, we get, 1.x [A, A(2A2 -1)] + y [A,A(2A2-1)] – (u + iv) = (1 +8i) (a + ib)2 2 30T4,A+2(SOA) = 0 (u – iv) = (1 -8i) (a – ib)2 2. R(A, 2A-1) – 101T4,A + G10A = 0

3. R (2A, 2A ) – 328 T4,A = 0 On equating real and imaginary parts, we have 4. x [1, A(A+1)]-7R[1,A(A+1)] – 18 PA –PT6 0 (mod 4) u = u (a, b) = a2 – b2 – 16ab v = v (a, b) = 8a2 – 8b2 + 2ab 5. y (A,1) + 9R(A,1) – 130 T4,A – G7A –PT1 = 0 2 2 2 12([.6 ), aayaax 12([]1 ),  OH a  0)(390]1 Substituting the values of u and v in (2), the non-zero distinct yx  )2(mod0)1,1()1,1(.7 values of x, y, z, w and R satisfying (1) are given by 2 2 2 2 aaax T  OH OH  x = x (a, b) = 9a – 9b – 14ab 12(,[.8 )] 585 ,4 a 21 a 390)( a 0 y = y (a, b) = - 7a2 + 7b2 – 18ab wz )1,1()1,1(.9  47 )3584(mod 4 4 2 2 3 3 z = z (a, b) = 2(8a + 8b – 48a b - 126a b + 126ab ) +1 2 6 10 aax 1([. ), 130]1  PaPa P  .0)(585 w = w (a, b) = 2 (8a4 + 8b4 – 48a2 b2 - 126a3b +126ab3) – 1 2 2 2 R = R (a, b) = a + b 2.4 Pattern: IV Rewrite (3) as Properties:- 1 * v2 = 65R2 – u2 (11)

2 2 2 Write 1 as 1 = (12) 1.7x [(2a-1) ,1] + 9y [(2a-1) ,1] + 260 (Ga) = 0 2 2 2. y (2a,1) + 7R (2a,1) + G9a – W1 0 (mod 2) Assume v = 65a – b = ( a - b) ( a + b) (13)

3. R (2a, 2a) – 8t4,a= 0 4. x (a,a+1) + y (a, a+1) + Ct 16,a + 24t4,a +G14a = 0 Using (12) and (13) in (11), it is written in factorizable form as

5. x (a,1) + R (a,1) – 10T4,a + G7a + TK2 0 (mod 2) 2 2 )2,2()2,2(.6 Twz 10,23  Carolnumber. ( a – b) ( a + b) 1([7.7 ), aayaax 1([9]1 ), P  .0288]1 a = ( R - u) ( R + u) (14) 2 2 2 ayax  Ga 0)(260])12(,1[9])12(,1[7.8 2 2 2 On equating the rational and irrational factors we get, 12(,1[9.9 )] aaxaay 12(,1[7 )] OH a )(378

192OH a T10 6.  0

10 yx )1,1(9)1,1(7. CS6  woodallnumber.

2.3 Pattern: III Replacing „a‟ by 8A and „b‟ by 8B in the above equations (13) and (15), we get 2 2 Rewrite (3) as R =R (A, B) = 520A + 8B + 16AB 2 2 1 * u2 = 65R2 – v2 (7) u = u (A, B) = 520A + 8B + 1040AB 2 2 Assume u =65a2 –b2 = ( (8) v = v (A, B) = 4160A – 64B

Write 1 a 1 = (9) On substituting the values of u and v in (2), the non –zero Using (8) and (9) in (7) it is written in factorizable form as distinct integrals values of x, y, z, w and R satisfying (1) are 2 2 ( a + b) ( a – b) given by = ( R +v) ( R –v) x = x (A, B) = 4680A2 – 56B2 + 1040AB 2 2 y = y (A, B) = – 3640A + 72B + 1040AB Volume 5 Issue 3, March 2016 www.ijsr.net Paper ID: NOV162379 Licensed Under Creative Commons Attribution CC BY 1864 International Journal of Science and Research (IJSR) ISSN (Online): 2319-7064 Index Copernicus Value (2013): 6.14 | Impact Factor (2014): 5.611

4 4 2 2 z = z (A, B) = 2 (2163200A – 512B + aax 12([.6 ), ]1  21 OH a )(  18OH a is cubica int eger 4326400A3B – 66560AB3) +1 Rzw PT  0)1,1()1,1()1,1(.7 w = w (A, B) = 2(2163200A4 – 512B4 + 6 3 3  Tay is perfecta square 4326400A B – 66560AB ) -1 7)1,1(.8 ,4 a . 2 2 R =R (A, B) = 520A + 8B + 16AB 2 2 aaR 12([.9 ),  OH a )(3]1  carol number.

10 9. [( 1), 2( )] 7 [( 1), aayaax 2( )]  292T Properties:- ,4 a G  1. x (A+1, 1)–y (A+1, 1 – 8320 T4,– G8320A-S5 (mod 2) 454a 2(mod0 ).

2. y (1A) – 9R (1,A) – G 0 (mod 3) 448A 3. Conclusion 3. R [A(2A2-1), 1] – 520 (SO )2 + 16 (SO ) –W = A A 2

Star number It is worth to note that in (2), the transformations for z and w 4. z (A, 1) – W(A,1) 0 (mod 2) may be considered as z = 2u + v and w = 2u –v. For this

5. x (1,B) + 7R (1,B) - P - G516 B 0 (mod 2) case, the values of x, y and R are the same as above where as the values of z and w changes for every pattern. To conclude R  41,1.6 is Nastya number. one may consider biquadratic equations with multivariables )1,1()1,1(.7 5 GnoPyx  0 25 34 (>5) and search for their non-zero distinct integer solutions 2 2 .8 72 aax 12(,[ )] 56 aay 12(,[ )] 133120T ,4 a along with their corresponding properties.

 99204 OH a 0 References ax 224)12,1(8.9 a 1040a 8  CullenJGP number. 10 wz )1,1()1,1(.  32(mod )802816 [1] Dickson, L.E., History of theory of numbers, Vol.11,

Chelsea publishing company, New –York (1952). 2.5 Pattern: V [2] Mordell, L.J.,Diophantine equation, Academic press,

London (1969) Journal of Science and Research,Vol (3) Write (3) as u2 - R2 = 64R2 - v2 Issue12, 20-22(December 14 (u + R) (u – R) = (8R + v) (8R – v), (16) [3] Jayakumar. P, Sangeetha, K “Lattice points on the cone which is expressed in the form of ratio as x2 + 9y2 =50z2” International Journal of Science and = , B  0 (17) Research, Vol (3), Issue 12, 20-22 (December -2014) [4] Jayakumar P, Kanaga Dhurga, C,” On Quadratic 2 2 2 This is equivalent to the following two equations Diophantine equation x +16y = 20z ” Galois J. Maths, - uA + R(8B +A) – VB = 0 1(1) (2014), 17-23. [5] Jayakumar. P, Kanaga Dhurga. C, “Lattice points on the uB + R (B – 8A) – VA = 0 2 2 2 cone x +9y =50z ” Diophantus J. Math, 3(2) (2014), On solving the above equations by the method of cross 61-71 [6] Jayakumar. P, Prabha. S “On Ternary Quadratic multiplication we get, 2 2 2 u = u (A, B) = –A2 – B2 –16AB Diophantine e quation x + 15y = 14 z ” Archimedes J. R = R (A, B) = – A2 – B2 Math., 4(3) (2014), 159-164. v = v (A, B) = 8A2 –8B2 – 2AB [7] Jayakumar, P, Meena, J “Integral solutions of the Ternary Quadratic Diophantine equation : x2 +7y2 = 2 Substituting the values of u and v in (2), the non – zero distinct 16z ” International Journal of Science and Technology, integral values of x, y, z, w and R satisfying (1) are given by, Vol.4, Issue 4, 1-4, Dec 2014. 2 2 [8] Jayakumar. P, Shankarakalidoss, G “Lattice points on x = x (A, B) = 7A – 9B – 18AB 2 2 2 y = y (A, B) = – 9A2 +7B2 –14AB Homogenous cone x +9y =50z ” International journal z = z (A, B) = 2[– 8A4 + 8B4 +32A2 B2 –126 A3B of Science and Research, Vol (4), Issue 1, 2053-2055, +126AB3] +1 January -2015. 4 4 2 2 [9] Jayakumar. P, Shankarakalidoss. G “Integral points on w = w (A, B) = 2[–8A +8B +32A B 2 2 2 126A3B+126AB3] – 1 the Homogenous cone x + y =10z International R = R (A, B) = - A2 - B2 Journal for Scienctific Research and Development, Vol (2), Issue 11, 234-235, January -2015 [10] Jayakumar.P, Prapha.S “Integral points on the cone x2 Properties : 2 2 2 +25y =17z ” International Journal of Science and 1.9x [1, A(A+1)] + 7y [1, A(A+1)] + 32(PA) – Research Vol(4), Issue 1, 2050-2052, January-2015. 260T4,A –G130A = woodall number [11] Jayakumar.P, Prabha. S, “Lattice points on the cone x2 + 2. R (A+1, 1) + T + G 0 (mod 3) 2 2 4,A A 9y = 26z “International Journal of Science and 2 2 3. y [ 1, A(2A -1)] + 7x [1,A(2A -1)] + 14 SOA + Research Vol (4), Issue 1, 2050-2052, January -2015 W3-Ky1 = 0 [12] Jayakumar. P, Sangeetha. K, “Integral solution of the 4. R (2A, 2A) + 8t4,A= 0 Homogeneous Biquadratic Diophantine equation with 5. x (1,1) + 7R (1,1) +P = Nasty number six unknowns: (x3-y3)z = (W2 – P2)R4 “International Journal of Science and Research, Vol (3), Issue 12, 1021-1023 (December-2014)

Volume 5 Issue 3, March 2016 www.ijsr.net Paper ID: NOV162379 Licensed Under Creative Commons Attribution CC BY 1865 International Journal of Science and Research (IJSR) ISSN (Online): 2319-7064 Index Copernicus Value (2013): 6.14 | Impact Factor (2014): 5.611 Author Profile

P. Jayakumar received the B. Sc, M.Sc degrees in Mathematics from Madras University in 1980 and 1983 and the M. Phil, Ph.D degrees in Mathematics from Bharathidasan University, Thiruchirappalli in 1988 and 2010.Who is now working as Professor of Mathematics, Periyar Maniammai University, Vallam, Thanajvur-613 403,Tamil Nadu,India.

R. Venkatraman received the B.Sc, M.Sc,and MPhil degrees in Mathematics from Bharathidasan University, Thiruchirappalli in 2002, 2004 and 2006. Who is now working as Assistant Professor of Mathematics, SRM University Vadapalani Campus, Chennai- 600026

Volume 5 Issue 3, March 2016 www.ijsr.net Paper ID: NOV162379 Licensed Under Creative Commons Attribution CC BY 1866