IJIRMPS | Volume 7, Issue 1, 2019 ISSN: 2349-7300 Integer solutions of non-homogeneous quintic equation with five unknowns 4 (x4 –y4) = 50 (z2-w2)m3 Dr. R. Anbuselvi Associate Professor, Department of Mathematics, A.D.M College for Women (Autonomous), Nagapattinam, India R. Nandhini Lecturer, Department of Mathematics, BDU Model College, Thiruthuraipoondi, India Abstract: The non – homogeneous quintic Diophantine equation with five unknowns 4 (x4 –y4) = 50 (z2-w2)m3 is considered and analysed for its non-zero distinct integer solutions. It is observed that by introducing the linear transformations x=u+v, y=u-v, z=8u+v, w=8u-v, (u ) in the given equation, one may obtain infinitely many non-zero integer solutions. Similarly one can introduce some other transformations to obtain different non-zero integer solutions. Keywords: Non-homogeneous quintic equation, quintic equation with five unknowns, polygonal number, pyramidal number, integer solutions. Introduction: The quintic Diophantine equation offers an unlimited field of research due to their variety [1-3]. For an extensive of various problems on quintic equation with three unknowns one may refer [4 & 5]. In [6-10] quintic equation with five unknowns are considered. This communication concern with another interesting quintic Diophantine equation with five unknowns 4 (x4 –y4) = 50 (z2-w2)m3 to find its integer solutions. Notations Used: tm,n - Polygonal number of rank ‘n’ with size ‘m’ SOn - Stella Octangular number of rank ‘n’ Obln - Oblong number of rank ‘n’ Jn - Jacobsthal number of rank ‘n’ jn - Jacobsthal – Lucas number of rank ‘n’ 5 Pa - Pentagonal pyramidal number n Pr - Pronic number of rank ‘n’ Kyn - Kynea number Caroln - Carol number Ictn - Icositetragonal number Method of Analysis: The non-homogeneous quintic equation with five unknowns under consideration is 4 (x4 –y4) = 50 (z2-w2)m3 (1) Pattern 1 : The substitution of the linear transformations x=u+v, y=u-v, z=8u+v, w=8u-v, (u ) (2) in (1) leads to u2 + v2 = 50 m3 (3) Assume m = m (a,b) = a2 + b2 (4) where a and b are non-zero distinct integers. Four different choices are presented here. Choice 1 : Write 50 as (1+7i) (1-7i) (5) (3) becomes u + iv = (1 + 7i) (a + ib)3 (6) Equating the real and imaginary parts on both sides, we have u = a3 - 21a2b - 3ab2 + 7b3 v = 7a3 - 3a2b - 21ab2 - b3 IJIRMPS1901029 Website : www.ijirmps.org Email : [email protected] 103 IJIRMPS | Volume 7, Issue 1, 2019 ISSN: 2349-7300 Employing the values of u and v in (2) we get x = x (a,b) = 8a3 - 18a2b - 24ab2 + 6b3 y = y (a,b) = -6a3 - 24a2b + 18ab2 + 8b3 z = z (a,b) = 15a3 - 165a2b - 45ab2 + 55b3 w =w(a,b) = a3 - 171a2b - 3ab2 + 57b3 m=m(a,b) = a2 + b2 Properties: 5 x (a,1) + y (a,1) - 4Pa + t64,a + t28,a 14 (mod 50) 9 y (1,b) + z(1,b) + w (1,b) -80CPb + t38,b + t26,b 10 (mod 348) 9 z (a,1) + m (a,1) – 10CPa + t250,a + 40 Pra 56 (mod 123) 14 x (2,b) + w (2,b) + m (2,b) – 27 CPb +t82,b + t28,b 176 (mod 771) n m (2 -1,4) -18 = caroln Choice 2: Write 50 as (7+i) (7-i) (7) (3) implies u + iv = (7+i) (a+ib)3 (8) By equating real and imaginary parts we get u = 7a3 - 3a2b - 21ab2 + b3 v = a3 + 21a2b - 3ab2 - 7b3 Substituting u and v in (2) we get x = x (a,b) = 8a3 + 18a2b - 24ab2 - 6b3 y = y (a,b) = 6a3 - 24a2b - 18ab2 + 8b3 z = z (a,b) = 57a3 - 3a2b - 171ab2 + b3 w = w(a,b) = 55a3 - 45a2b - 165ab2 + 15b3 m = m(a,b) = a2 + b2 Properties : 9 . z (a,2) – m (a,2)-38CPa + t10,a + t8,a 4 (mod 670) 5 . x(a,a+1)+8Pa + t42,a+t50,a 6 (mod 84) 3 . z(a,1)-x(a,1)+m(a,1)-98CPa + t26,a +t18,a 8 (mod 214) 14 . w(a,2) – y(a,2) – 21CPa + t62,a + t26,a 16 (mod 40) n . m (2 ,5) = 3 J2n + 26 Choice 3: 50 can be written as 50 = (5 + 5i) (5 – 5i) (9) By using (4) and (9) in (3) and applying the same procedure as mentioned above we get the corresponding integer solutions to (1) as x = x (a,b) = 10a3 – 30ab2 y = y (a,b) = 10b3 – 30a2b z = z (a,b) = 45a3 – 105a2b – 135ab2 + 35b3 w = w (a,b) = 35a3 – 135a2b – 105ab2 + 45b3 m = m (a,b) = a2 + b2 Choice 4 : Instead of (9), 50 can also be written as 50 = (10 + 10i) (10 – 10i) (10) 4 By using (4) and (10) in (3) and proceeding as above, the corresponding integer solutions to (1) are given below. x = x(a,b) = ½ (20a3 – 60ab2) y = y(a,b) = ½ (20b3 – 60a2b) IJIRMPS1901029 Website : www.ijirmps.org Email : [email protected] 104 IJIRMPS | Volume 7, Issue 1, 2019 ISSN: 2349-7300 z = z (a,b) = ½ (90a3 – 270ab2 – 210a2b + 70b3 ) w = w (a,b) = ½ (70a3 – 210ab2 – 270a2b + 90b3) m = m (a,b) = a2 + b2 It is worth to mention here that 50 can also be represented as follows 50 = (15 + 15i) (15 – 15i) 9 50 = (20 + 20i) (20 – 20i) 16 50 = (25 + 25i) (25 – 25i) 25 and so on Pattern 2 : We can write (3) as u2 + v2 = 50 m3 * 1 (11) Write 1 as 1 = (4 + 3i) (4 – 3i) (12) 25 Using (4), (5) and (12) in (11) and applying the method of factorization, we may define. u + iv = 4+3i (1 + 7i) (a + ib)3 5 Equating real and imaginary parts we get u = 1/5 (-17a3 + 51ab2 – 93a2b + 31b3 ) (13) v = 1/5 (31a3 – 93ab2 – 51ab2 + 17b3 ) As our aim is to find integer solutions, we may choose a = 5A, b = 5B in (4) and (13) and substituting the values of u and v in (2), we obtain the non-zero integer solutions to (1) as x = 25(14A3 – 42AB2 – 144A2B + 48B3) y = 25(- 48A3 + 144AB2 – 42A2B + 14B3) z = 25 (-105A3 + 315AB2 – 795A2B + 265B3) w = 25 (-167A3 + 501AB2 – 693A2B + 231B3) m = 25 (A2 + B2) Properties : 9 3 y(1,2B) + m(1,2B) - 1400CPB + 1400CPB – 700SO3 – t28002,B– t1002,B 1225 (mod 14498) 6 z(2,B) – 7950CPB – 2ICtn – t29958,B –t1502,B -5274 (mod 15726) n m (2 +1, 1) – 75 = 25 kyn m (A, A+1) – 25 = 50 OblB x (1,B) – 600SOB + t2002, B + t102, B 350 (mod 4048) Pattern 3: By using different set of transformation as x = u + v, y = u – v, z = 4u + 4v, w = 4u – 4v, u v 0 (14) in (1), we get u2 + v2 = 100 m3 (15) Choice 1 : Write 100 as 100 = (8 + 6i) (8 – 6i) (16) By applying the same process as mentioned in the previous choices, we get integer solutions to (1) using (14) as x = x (a,b) = 14a3 – 42ab2 + 6a2b – 2b3 y = y (a,b) = 2a3 – 6ab2 – 42a2b + 14b3 z = z (a,b) = 56a3 – 168ab2 + 24a2b - 8b3 w = w (a,b) = 8a3 – 24ab2 – 168a2b + 56b3 IJIRMPS1901029 Website : www.ijirmps.org Email : [email protected] 105 IJIRMPS | Volume 7, Issue 1, 2019 ISSN: 2349-7300 m = m (a,b) = a2 + b2 Properties : n n+1 . m(2 , 2 ) = 3J2n + j2n+2 14 . z (a,2) – 24CPa – t42,a – t58,a -64 (mod 594) 5 . x (a,3) – y(a,3) – 24Pa – t202,a –t66,a - 44 (mod 194) 3 . y (1,b) + w (1,b) – 140CPb + t38,b + t26,b 10 (mod 308) 5 6 . w (a,a+1) + 336Pa – 48CPa – t162,a – t82,a 56 (mod 254) Choice 2 : Instead of (15), we can write 100 as 100 = (6+8i) (6-8i) (17) By applying the same procedure we mentioned above, we obtain the following integer solutions to (1) x = x (a,b) = 14a3 – 42ab2 – 6a2b + 2b3 y = y (a,b) = - 2a3 + 6ab2 – 42a2b + 14b3 z = z (a,b) = 56a3 – 168ab2 – 24a2b + 8b3 w = w (a,b) = - 8a3 + 24ab2 – 168a2b + 56b3 m = m (a,b) = a2 + b2 Choice 3: In the place of (15), we can put 100 as 100 = (16 + 12i) (16 – 12i) (18) 4 Substituting (4) and (18) in (15) and applying the method of factorization, we can define u + iv = ½ (16 + 12i) (a + ib)3 Equating real and imaginary parts we get u = ½ (16a3 – 48ab2 – 36a2b + 12b3) (19) v = ½ (12a3 – 36ab2 + 48a2b – 16b3) Assuming a = 2A and b = 2B in (19) u = 4 (16A3 – 48AB2 – 36A2B + 12B3) v = 4 (12A3 – 36AB2 + 48A2B – 16B3) From (14), the integer solutions to (1) are x = x(A,B) = 4 (28A3 – 84AB2 + 12A2B – 4B3) y = y(A,B) = 4 (4A3 – 12AB2 – 84A2B + 28B3) z = z(A,B) = 16 (28A3 – 84AB2 + 12A2B – 4B3) w = w(A,B) = 16 (4A3 – 12AB2 – 84A2B + 28B3) m = m (A,B) = 4 (A2 + B2) It is worth to mention here that 100 can also be represented as follows 100 = (24 + 18i) (24 – 18i) 9 100 = (32 + 24i) (32 – 24i) 16 100 = (40 + 30i) (40 – 30i) 25 and so on Conclusion: In this paper, we have taken two different set of transformations and some patterns to obtain distinct integer solutions. To conclude, one may try some other set of transformations and patterns to find different integer solutions. IJIRMPS1901029 Website : www.ijirmps.org Email : [email protected] 106 IJIRMPS | Volume 7, Issue 1, 2019 ISSN: 2349-7300 References: 1. Carmicheal R.D “The Theory of numbers and Diophantine Analysis, Dover publications, New York (1959). 2. Mordell.L.J, “Diophantine Equations” Academic press, New York, 1969. 3. Dickson.L.E., History of theory of Numbers, Vol.2, Chelsia Publishing Co., New York (1952) 4.