Solutions of the diophantine equation 2 x + p y = z 2

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1 Int. J. of Mathematical Sciences and Applications, Vol. 1, No. 3, September 2011 Copyright Mind Reader Publications Solutions of the diophantine equation 2 x + p y = z 2 Alongkot Suvarnamani Department of Mathematics, Faculty of Science and Technology, Rajamangala University of Technology Thanyaburi (RMUTT), Thanyaburi, Pathum Thani, 12110, Thailand. kotmaster2@rmutt.ac.th kotmaster2@hotmail.com Abstract In this paper, we study the diophantine equation 2 x + p y = z 2 where where p is a prime number and x, y and z are non-negative integers. 1 Introduction In 2002, J. Sand studied two diophantine equations 3 x + 3 y = 6 z and 4 x + 18 y = 22 z. After that D. Acu (2007) studied the diophantine equation of fm 2 x + 5 y = z 2. He found that this equation has exactly two solutions in non-negative integer (x, y, z) {(3, 0, 3), (2, 1, 3)}. Then A. Suvarnamani, A. Singta and S. Chotchaisthit (2011) found solutions of two diophantine equations 4 x + 7 y = z 2 and 4 x + 11 y = z 2. Now, we study the diophantine equation of fm 2 x + p y = z 2 (1), where p is a prime number and x, y and z are non-negative integers. 2 Main Theem From the diophantine equation (1), we have 2 x + 2 y = z 2 (2), where p = 2. From the diophantine equation (2), we consider in 3 cases Mathematics Subject Classification: 11D61. Key wds and phrases: diophantine equations, exponential equations

2 Alongkot Suvarnamani 2 A. Suvarnamani case 1: x = y. The diophantine equation (2) becomes 2 x+1 = z 2. So, z = 2 k where k is a non-negative integer. That is x = 2k 1. But it is impossible f k = 0. Hence, the solution of the diophantine equation (2) is (x, y, z) = (2k 1, 2k 1, 2 k ) where k is a positive integer. case 2: x > y. The diophantine equation (2) becomes 2 y (2 x y + 1) = z 2. That is z = a 2 k where k is a non-negative integer and a 2 = 2 x y + 1. So, y = 2k. Since a 2 = 2 x y + 1, we get 2 x y = a 2 1 = (a 1)(a + 1) = 2 v 2 x y v, where a 1 = 2 v and a + 1 = 2 x y v, x y > 2v and v is a non-negative integer. Then we obtain 2 v (2 x y 2v 1) = 2. F v = 0, we get 2 x y 1 = 2. It is impossible. F v = 1, we have 2 x y 2 1 = 1. So, 2 x y 2 = 2. That is x y 2 = 1. Then we get x = 2k + 3 and a = 3. Hence, the solution of the diophantine equation (1) is (x, y, z) = (2k + 3, 2k, 3 2 k ) where k is a non-negative integer. case 3: x < y. It is similarly with case 2. Hence, the solution of the diophantine equation (2) is (x, y, z) = (2k, 2k + 3, 3 2 k ) where k is a non-negative integer. In conclusion, we have Main Theem 2.1. Let A = {(2k 1, 2k 1, 2 k ) k is a positive integer.}, B = {(2k + 3, 2k, 3 2 k ) k is a non-negative integer.} and C = {(2k, 2k + 3, 3 2 k ) k is a non-negative integer.}. The solution of the diophantine equation (2) is (x, y, z) A B C. 1416

3 Solutions of the diophantine equation Solutions of the diophantine equation 2 x + p y = z 2 3 Main Theem 2.2. Consider the diophantine equation (1) where p is a prime number which is me than 2. (i) F each prime number p, the diophantine equation (1) has a solution (x, y, z) = (3, 0, 3). (ii) F p = 3, the diophantine equation (1) has a solution (x, y, z) = (4, 2, 5). (iii) F p = k+1 where k is non-negative integer, the diophantine equation (1) has a solution (x, y, z) = (2k, 1, k ). Proof. From the diophantine equation (1), we consider in 2 case. Case 1: if x is an odd number. That is x = 2k + 1 where k is a non-negative integer. The diophantine equation (1) becomes z 2 2 2k+1 = p y (z 2 k+ 1 2 )(z + 2 k+ 1 2 ) = p y, where z 2 k+ 1 2 = p u and z + 2 k+ 1 2 = p y u, y > 2u and u is a non-negative integer. Then we obtain p y u p u = 2 k+ 3 2 p u (p y 2u 1) = 2 k Clearly, it is impossible if u > 0. F u = 0, we get p y 1 = 2 k+ 3 2 p y 2 k+ 3 2 = 1 (3). The diophantine equation (3) is a diophantine equation by Catalan s type a b c d = 1. So, it has in positive integer number (> 1) only the solutions p = 3, y = 2 and k = 3. But it is impossible. F y = 1, we get p 2 k+ 3 2 = 1. It is impossible, too. F y = 0, we get z 2 2 2k+1 = 1 which it is no solution if z = 0 z = 1. However, it is a diophantine equation by Catalan s type a b c d = 1. So, it has in positive integer number(> 1) only the solutions z = 3 and 2k + 1 = 3. That is k = 1. Hence, a 1417

4 Alongkot Suvarnamani 4 A. Suvarnamani solution of the diophantine equation (1) is (x, y, z) = (3, 0, 3). Case 2: if x is an even number. That is x = 2k where k is a non-negative integer. The diophantine equation (1) becomes z 2 2 2k = p y (z 2 k )(z + 2 k ) = p y, where z 2 k = p v and z + 2 k = p y v, y > 2v and v is non-negative integer. Then we obtain p y v p v if v > 0. = 2 k+1 p v (p y 2v 1) = 2 k+1. Clearly, it is impossible F v = 0, we get p y 1 = 2 k+1 p y 2 k+1 = 1 (4). The diophantine equation (4) is a diophantine equation by Catalan s type a b c d = 1. So, it has in positive integer number (> 1) only the solutions p = 3, y = 2 and k + 1 = 3. That is k = 2. Hence, a solution of the diophantine equation (1) is (x, y, z) = (4, 2, 5) if p = 3. F y = 1, we get p 2 k+1 = 1 p = k+1. Hence, a solution of the diophantine equation (1) is (x, y, z) = (2k, 1, k ) if p = k+1. F y = 0, we get z 2 2 2k = 1. It is impossible. Acknowledgements I would like to thank the referee(s) f his comments and suggestions on the manuscript. This wk was suppted by the Faculty of Sciences and Technology, Rajamangala University of Technology Thanyaburi (RMUTT), Thailand. 1418

5 Solutions of the diophantine equation Solutions of the diophantine equation 2 x + p y = z 2 5 References [1] D. Acu, On a diophantine equation 2 x + 5 y = z 2, General Mathematics, Vol. 15, No. 4 (2007), [2] M.B. David, Elementary Number They, 6th ed., McGraw-Hill, Singape, [3] H.R. Kenneth, Elementary Number They and its Application, 4th ed., Addison Wesley Longman, Inc., [4] L.J. Mdell, Diophantine Equations, Academic Press, New Yk, [5] J. Sand, On a diophantine equation 3 x + 3 y = 6 z, Geometric theems, Diophantine equations, and arithmetric functions, American Research Press Rehobot 4, 2002, [6] J. Sand, On a diophantine equation 4 x + 18 y = 22 z, Geometric theems, Diophantine equations, and arithmetric functions, American Research Press Rehobot 4, 2002, [7] W. Sierpinski, Elementary They of Numbers, Warszawa, [8] J.H. Silverman, A Friendly Introduction to Number They, 2nd ed., Prentice-Hall, Inc., New Jersey, [9] A. Suvarnamani, A. Singta and S. Chotchaisthit On two Diophantine equations 4 x + 7 y = z 2 and 4 x + 11 y = z 2, Science and Technology RMUTT Journal, Vol. 1, No.1 (2011). 1419