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Use Style: Paper Title DOI 10.4010/2016.1309 ISSN 2321 3361 © 2016 IJESC Research Article Volume 6 Issue No. 5 Integral Solutions of the Non-Homogeneous Heptic Equation with Five Unknowns 6(x2 y2 ) 11xy 3x 3y 9 72z2 G.Janaki1, C.Saranya2 Assistant Professors1, 2 Department of Mathematics Cauvery College for Women, Tiruchirappalli, Tamil Nadu, India [email protected], [email protected] Abstract: The non-homogeneous Diophantine equation of degree seven with five unknowns represented by 5(x3 y3) 7(x2 y2 ) 4(z3 w3 3wz xy 1) 972 p7 is analyzed for its non-zero distinct integer solutions. A few interesting relations between the solutions and some special numbers are presented. Keywords: Non-homogeneous heptic, heptic equation with five unknowns, Integral solutions. 2010 Mathematics Subject Classification: 11D 41 I. INTRODUCTION j = Jacobsthal-Lucas number of rank ‘n’. Mathematics is the language of patterns and relationships and n TT = Truncated tetrahedral number of rank ‘n’. is used to describe anything that can be quantified. n Diophantine equations has been matter of interest to various CH = Centered Hexagonal number of rank ‘n’. mathematicians. The problem of finding all integer solutions n of a diophantine equation with three or more variables and J = Jacobsthal number of rank ‘n’. n equations of degree at least three, in general presents a good deal of difficulties. A lot is known about equations in two III. METHOD OF ANALYSIS variables in higher degrees. There is vast general theory of The heptic equation to be solved for its non-zero integral homogeneous quadratic equations with three variables. In [1- solution is 3], theory of numbers were discussed. In [4,5], a special (1) Pythagorean triangle problem have been discussed for its integral solutions. In [6-10], higher order equations are On substitution of the transformations, considered for integral solutions. x u 1, y u 1, z v 1,w v 1 (2) In this communication a seventh degree non-homogeneous in (1) leads to, equation, with five variables represented by 2 2 7 u 3 v 81 p (3) We illustrate below four different patterns of non-zero distinct is considered and in particular a few interesting relations integer solutions to (1). among the solutions are presented. Pattern: 1 II. NOTATIONS Assume p p (a,b) a2 3 b2 (4) Obl = Oblong number of rank ‘n’. n where a and b are non-zero integers. T = Polygonal number of rank ‘n’ with sides ‘m’. Substituting (4) in (3), and using factorization method, m,n (u i 3v) (u i 3v) 81 (a i 3b)7 (a i 3b)7 (5) CSn = Centered Square number of rank ‘n’. Equating the like terms and comparing real and imaginary SOn = Stella octangula number of rank ‘n’. parts, we get 7 5 2 3 4 6 On = Octahedral number of rank ‘n’. u u (a,b) 9a 567 a b 2835 a b 1701 a b Gno = Gnomonic number of rank ‘n’. 6 4 3 2 5 7 n v v (a,b) 63 a b 945 a b 1701 a b 243 b Star = Star number of rank ‘n’. n Tha = Thabit-ibn-kurrah number of rank ‘n’. Substituting the above values of u &v in equation (2), the n corresponding integer solutions of (1) are given by Carln = Carol number of rank ‘n’. K n = Kynea number of rank ‘n’. International Journal of Engineering Science and Computing, May 2016 5347 http://ijesc.org/ 7 5 2 3 4 6 Properties: x x (a,b) 9a 567 a b 2835 a b 1701 a b 1 7 5 2 3 4 6 1. y(A, A) x (A, A) p (A, A) StarA T8, A T12, A y y (a,b) 9a 567 a b 2835 a b 1701 a b 1 6 4 3 2 5 7 2 Obl 5 Gno 0 (mod 6) z z (a,b) 63 a b 945 a b 1701 a b 243 b 1 (6) A A 2. z (A, 1) w (A, 1) p (A, 1) 3O SO 2 CS Gno 0 (mod 13) 6 4 3 2 5 7 A A A A w w (a,b) 63 a b 945 a b 1701 a b 243 b 1 2 2 3. y (A, 1) x (A, 1) p (A, 1) CH 2 T Gno 0 (mod 14) p p (a,b) a 3 b A 3, A A n n n Properties: 4. x (2 , 1) y (2 , 1) p (2 , 1) 4K n 4 jn 12 J n 0 (mod 18) 1. z (a,a) w (a,a) p (a,a) CS a 4 T3, a 0 (mod 3) 5. z (A, A) w (A, A) p (A, A) T30, A 2 OblA T10, A 9 Gno 0 (mod 7) 2. x (a,a) y (a,a) p (a,a) CS a 2 Obla 0 (mod 3) A Pattern: 3 3. x (2n , 1) y (2n , 1) 2p (2n , 1) 2K 2 j n n (1 4i 3) (1 4i 3) 6 J 0 (mod 10) Instead of (7), Write 1 (8) n 49 n 4. 3p (2 , 1) 2 Than 3 Carln 0 (mod 14) Substituting (8) and (4) in (6) and employing the method of 5. x (a,a) y (a,a) p (a,a) Star T factorization, following the procedure presented in pattern 2, a 6, a the corresponding integer solutions of (1) are represented by 7 6 5 2 4 3 T14, a 6 Obla 3 Gnoa 0 (mod 2) A 84 A B 63 A B 1260 A B x x (A, B) 32. 76 1 3 4 2 5 6 7 315 A B 2268 A B 189 AB 324 B Pattern: 2 A7 84 A6B 63 A5B2 1260 A4B3 The equation (3) can be written as y y (A, B) 32. 76 1 3 4 2 5 6 7 2 2 7 315 A B 2268 A B 189 AB 324 B u 3 v 81 p 1 (6) 4 A7 7 A6B 252 A5B2 105 A4B3 z z (A, B) 32. 76 1 (9) (1 i 3) (1 i 3) 3 4 2 5 6 7 and write 1 1260 A B 189 A B 756 AB 27 B (7) 4 7 6 5 2 4 3 2 6 4 A 7 A B 252 A B 105 A B Substituting (4) and (7) in (6), and using factorization w w (A, B) 3 . 7 1 1260 A3B4 189 A2B5 756 AB 6 27 B7 method, 2 2 (1 i 3) (1 i 3) p p (A, B) 49 A 147 B 7 7 (u i 3v) (u i 3v) 81 (a i 3b) (a i 3b) 4 Equating the like terms and comparing real and imaginary Properties: parts, we get 1. y(A, A) x (A, A) p (A, A) 32 StarA 4 OblA 9 a7 21 a6b 63 a5b2 315 a 4b3 315 a3b4 u u (a,b) 94 GnoA 0 (mod 64) 2 2 5 6 7 567 a b 189 ab 81 b 2. z (A, A) w (A, A) p (A, A) 64CH A 9 a7 7 a6b 63 a5b2 105 a 4b3 315 a3b4 v v (a,b) 2 CS A 98GnoA 0 (mod 30) 2 5 6 7 2 189 a b 189 ab 27 b 3. 2 6 TTA p (A, 1) 2 T28, A 51 GnoA 0 (mod 325) n n n Since our interest is on finding integer solutions, we have 4. x (2 , 1) y (2 , 1) p (2 , 1) choose and suitably so that and are integers. Let us a b u v 49 ( jn 3 J n K n ) 0 (mod 198) take a 2A and b 2B , n n n 5. z (2 , 1) w (2 , 1) 3p (2 , 1) 98 Than In view of (2), the integer solutions of (1) are given by 147 Carln 0 (mod 684) Pattern: 4 A7 21 A6 B 63 A5B2 315 A4B3 315 A3B4 x x (A, B) 32. 26 1 Instead of (7), write 2 5 6 7 567 A B 189 AB 81 B A7 21 A6 B 63 A5B2 315 A4 B3 315 A3B4 y y (A, B) 32. 26 1 (13 3i 3) (13 3i 3) 567 A2 B5 189 AB 6 81 B7 1 (10) 196 A7 7 A6B 63 A5B2 105 A4B3 315 A3B4 z z (A, B) 32. 26 1 2 5 6 7 Substituting (9) and (4) in (6) and employing the method of 189 A B 189 AB 27 B factorization, following the procedure presented in pattern 2, A7 7 A6B 63 A5B2 105 A4B3 315 A3B4 w w (A, B) 32. 26 1 the corresponding integer solutions of (1) are represented by 2 5 6 7 189 A B 189 AB 27 B p p (A, B) 4 A2 12 B2 International Journal of Engineering Science and Computing, May 2016 5348 http://ijesc.org/ 13A7 63 A6 B 819 A5B2 945 A4 B3 V. REFERENCES x x (A, B) 32. 146 1 3 4 2 5 6 7 [1] Carmichael, R.D., The theory of numbers and 4095 A B 1701 A B 2457 AB 243 B Diophantine Analysis, Dover Publications,New York, 13A7 63 A6 B 819 A5B2 945 A4 B3 y y (A, B) 32. 146 1 1959. 4095 A3B4 1701 A2 B5 2457 AB 6 243 B7 3 A7 91 A6 B 189 A5B2 1365 A4 B3 [2] Dickson L.E, History of Theory of Numbers, Vol.11, z z (A, B) 32.
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