Solution of the diophantine equation 4x + py = z2n

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Selçuk J. Appl. Math. Selçuk Journal of Vol. 13. No. 2. pp. 31-34, 2012 Applied Mathematics

Solution of the Diophantine Equation4x_{+ p}y_{= z}2n Selin (Inag) Cenberci, Bilge Peker

Secondary Mathematics Education Programme, Ahmet Kelesoglu Education Faculty, Necmettin Erbakan University, Konya, Turkiye

e-mail:inag_ s@ hotm ail.com

Elementary Mathematics Education Programme, Ahmet Kelesoglu Education Faculty, Necmettin Erbakan University, Konya, Turkiye

e-mail:bilge.p eker@ yaho o.com

Received Date: February 9, 2012 Accepted Date: December 6, 2012

Abstract. In this paper, we gave solution of the Diophantine equation 4x₊ py _{= z}4 _{when p is an odd prime and we considered solution of the Diophantine} equation 4x_{+ p}y _{= z}2n _{where p > 2 is a prime number, n > 2 and x; y; z are} non-negative integers.

Key words: Exponential Diophantine Equation. 2000 Mathematics Subject Classi…cation: 11D61. 1. Introduction and Preliminaries

Solutions of the Diophantine equation of type ax+ by = cz attracted mathe-maticians for many years: In [3], Cao showed that this equation have at most one solution with z > 1: In [2], Acu found that the Diophantine equation 2x_{+ 5}y _{= z}2 _{has exactly two solutions in non-negative integers, i.e. (x; y; z) 2} f(3; 0; 3) ; (2; 1; 3)g : In [12] ; Suvarnamani considered the Diophantine equation 2x_{+ p}y _{= z}2 _{where p is a prime and x; y; z are non-negative integers.} Suvar-namani, Singta and Chotchaisthit found solutions of the Diophantine equations 4x_{+ 7}y _{= z}2 _{and 4}x_{+ 11}y _{= z}2 _{in [13] : Sandor studied on the Diophantine} equation 4x_{+ 18}y _{= 22}z _{[9] : Selberg proved that there is no positive integer} solution of the equation x4 _{1 = y}n _{for n} _{2 in [6; 8]. In [4] ;Chotchaisthit} showed all non-negative integer solutions of 4x_{+ p}y _{= z}2_{, where p is a prime} number. Peker and Cenberci studied on the solutions of the Diophantine equa-tion of some types of (2n₎x_{+ p}y _{= z}2 _{where p is an odd prime and x; y; z are} non-negative integers [7] :

In this study, we gave solution of the Diophantine equations 4x+ py = z4 where p is an odd prime and we considered a generalization for the equation

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4x+py= z2nwhere p > 2 is a prime, n > 2 and x; y; z are non-negative integers. While we are doing these, we use the Catalan’s Conjecture [5] and a proposition. Now we give them.

Conjecture 1. (Catalan) The only solution in integers a > 1; b > 1; x > 1 and y > 1 of the equation ax _by_{= 1 is a = y = 3 and b = x = 2:}

Proposition 1. [8,9] The equation x4 _{1 = y}n _{has no positive integer solution} for n 2:

2. Main Results and its Consequences Now we give our theorems.

Theorem 1. The Diophantine equation

(1) 4x+ py= z4

has only one non-negative integer solution, namely (x; y; z; p) = (3; 1; 3; 17) where p > 2 is a prime number and x; y; z are non-negative integers.

Proof. If we consider y > 0, then we get z4 22x= py i.e.

(z2 ₂x_)(z2_{+ 2}x_{) = p}y

where z2 2x= pvand z2+ 2x= py v; y > 2v and v is a non-negative integer. Then we get py v _pv_{= 2}x+1 _{or p}v_(py 2v _{1) = 2}x+1_:

If v = 0, we obtain py _{1 = 2}x+1 _{or p}y ₂x+1 _{= 1. From the Catalan}0_s Conjecture, it is obvious that p = 3; y = 2 and x + 1 = 3; x = 2: If we write these values in the equation (1), we can not …nd a z 2 Z+_{. So the equation (1)} has no non-negative integer solution.

For y = 1, we have p 2x+1 _{= 1 , i.e. p = 1 + 2}x+1_{. If we write this in the} equation (1), then we obtain z2 ₂x _{= 1: From the Catalan}0_{s Conjecture,} the only solution is given by z = 3; x = 3 and consequently a solution of the equation (1) is (x; y; z; p) = (3; 1; 3; 17).

For y = 0, we get z4 ₂2x_{= 1; i.e. z}4 _{1 = 2}2x_{which has no solution for x} ₁ from P roposition 1. For x = 0; we have z4 _{1 = 1 which is impossible.} In the Diophantine equation (1) ; if we consider x = 0, then the equation be-comes

1 + py = z4 i.e.

z4 1 = py:

This equation gives us P roposition 1. So we can say that this equation has no solution for y 2: If we consider y = 1; then we get p = z4 1 =

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(z 1) (z + 1) z2+ 1 which is impossible since p is a prime. For y = 0; we …nd z4 1 = 1; i.e. z4= 2 which is impossible. This completes the proof. Theorem 2. The Diophantine equation

(2) 4x+ py = z2n

has no non-negative integer solution where p > 2 is a prime number, x; y; z are non-negative integers and n > 2 is an integer:

Proof. If we consider y > 0, then we get z2n 22x= py i.e.

(zn 2x)(zn+ 2x) = py

where zn 2x= pv and zn+ 2x= py v; y > 2v and v is a non-negative integer. Then we get py v _pv_{= 2}x+1 _{or p}v_(py 2v _{1) = 2}x+1_:

If v = 0, we obtain py _{1 = 2}x+1 _{or p}y ₂x+1 _{= 1. From the Catalan}0_s Conjecture, it is obvious that p = 3; y = 2 and x + 1 = 3; x = 2: If we write these values in the equation (2), then we get 42_{+ 3}2_{= z}2n_{. This equation has} only one solution when n = 1: The solution of this equation is (x; y; z; p) = (2; 2; 5; 3) given by Chotchaisthit [4] :

For y = 1, we obtain p 2x+1 _{= 1, i.e. p = 1 + 2}x+1_{. If we write this in} the equation (2), then we obtain zn ₂x_{= 1: For n = 1; we …nd 2}x_{+ 1 = z .} Therefore the solutions of the equation (2) is (x; y; z; p) = k; 1; 2x_{+ 1; 2}x+1_{+ 1} given by Chotchaisthit [4] : For n = 2; we …nd 2x_{+ 1 = z}2_{; i.e. 1 = z}2 ₂x_: From the Catalan0_{s Conjecture, we get z = 3; x = 3 and p = 17. So, the only} solution of the equation (2) is (x; y; z; p) = (3; 1; 3; 17) obtained in T heorem 1. For n > 2; we obtain 2x+ 1 = zn; i.e. 1 = zn 2x. From the Catalan0s Conjecture, we do not …nd a solution for n > 2:

For y = 0, we get z2n 22x = 1; which has no solution from the Catalan0s Conjecture too.

In the Diophantine equation (2) ; if we consider x = 0, then the equation be-comes

1 + py= z2n i.e.

z2n py= 1:

From the Catalan0_{s Conjecture solution of this equation is only (z; y; p; n) =} (3; 3; 2; 1) : However, p is even. This contradicts to our assumption that p is an odd prime. So we can say that this equation has no non-negative integer solution. This completes the proof.

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