Non-Negative Integer Solutions of Two Diophantine Equations 2^x + 9^y = z² and 5^x + 9^y = z²

Md. Al-Amin Khan, Abdur Rashid, Md. Sharif Uddin

Department of Mathematics, Jahangirnagar University, Dhaka, Bangladesh.

DOI: 10.4236/jamp.2016.44086                

Abstract

In this paper, we study two Diophantine equations of the type p^x + 9^y = z² , where p is a prime number. We find that the equation 2^x + 9^y = z² has exactly two solutions (x, y, z) in non-negative integer i.e., {(3, 0, 3),(4, 1, 5)} but 5^x + 9^y = z² has no non-negative integer solution.

Keywords

Exponential Diophantine Equation, Integer Solutions

Received 7 October 2015; accepted 24 April 2016; published 27 April 2016

1. Introduction

Recently, there have been a lot of studies about the Diophantine equation of the type. In 2012, B. Sroysang [1] proved that is a unique solution for the Diophantine equation where and z are non-negative integers. In 2013, B. Sroysang [2] showed that the Diophantine equation has a unique non-negative integer solution. In the same year, B. Sroysang [3] found all the solutions to the Diophantine equation where and z are non-negative integers. The solutions are, and. In 2013, Rabago [4] showed that the solutions of the two Diophantine equations and where and z are non-nega- tive integers are and, respectively. Different examples of Diophantine equations have been studied (see for instance [5] - [11] ).

In this study, we consider the Diophantine equation of the type where p is prime. Particularly, we show that has exactly two solutions in non-negative integer and has no non-negative integer solution.

2. Main Results

Theorem 2.1. (Catalan’s Conjecture [12] ) The Diophantine equation, where and y are integers with, has a unique solution.

Theorem 2.2. The Diophantine equation has a unique non-negative integer solution.

Proof: Let x and z be non-negative integers such that. For, which is impossible. Suppose. Then,. Let and, where. Thus, or,. Now we have two possibilities.

Case-1: If, then. These give us and. Then and. Thus is a solution of.

Case-2: If, then. These give us and which is impossible.

Hence, is a unique non-negative integer solution for the equation.

Theorem 2.3. The Diophantine equation, where p is an odd prime number, has exactly one non-negative integers solution.

Proof: Let x and z be non-negative integers such that, where p be an odd prime. If, then. It is impossible. If, then, which is also impossible. Now for,

or.

Let and, where,. Then,

or.

Thus, and, which is possible only for and. So,.

Therefore, is the solution of. This proves the theorem.

Corollary 2.4. The Diophantine equation has no non-negative integers solution.

Theorem 2.5. The Diophantine equation has no unique non-negative integer solution.

Proof: Suppose x and z be non-negative integers such that. For, we have. It is impossible. Let. Then gives us. Let and, where,. Therefore,

or.

Thus, or and or. So, which is not acceptable since x is a non-negative integer. This completes the proof.

Theorem 2.6. The Diophantine equation has exactly two solutions in non-negative integer i.e.,.

Proof: Suppose and z are non-negative integers for which. If, we have which has no solution by theorem 2.5. For, by theorem 2.2 we have and. Hence is a solution to. If, then which is not possible for any non-negative integers x and y.

Now we consider the following remaining cases.

Case-1:. If, then or. We have two possibilities. If

and, then or but which is not acceptable. On the other hand, if and

same thing is occurred.

Case-2:. If, then or. Let and, where. Then or. Thus, and, then this implies that and or. So and. Here we obtain the solution.

Case-3:. If, then which is not possible for any for any non-negative integers x and y.

Case-4:. Now

or.

Let and, where. So or . Thus, and then these imply that and. So we get

(1)

The Diophantine Equation (1) is a Diophantine equation by Catalan’s type because for, the value of must be grater that 1. So by the Catalan’s conjecture Equation (1) has no solution. This proves the theorem.

Theorem 2.7. The Diophantine equation has no non-negative integer solution.

Proof: Suppose and z are non-negative integers for which. If, we have which has no solution by Theorem 2.5. For we use corollary 2.4. If, then which is not possible for any non-negative integers x and y.

Now we consider the following remaining cases.

Case-1:. If, then or. We have two possibilities. If and, it follows that or and , a contradiction. On the other hand, and, it follows that or and which is impossible.

Case-2:. If, then or. Let and, where. Then or. Thus, and, then this implies that and, a contradiction.

Case-3:. If, then which is not possible for any for any non-negative integers x and y.

Case-4:. Now

or

Let and, where. So or . Thus, and then these imply that and. Since, it follows that i.e.,. But we see that. This is impossible.

3. Conclusion

In the paper, we have discussed two Diophantine equation of the type, where p is a prime number. We have found that and are the exact solutions to in non-negative integers. On the contrary, we have also found that the Diophantine equation has no non-negative integer solution.

Conflicts of Interest

The authors declare no conflicts of interest.

References

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