On the exponential Diophantine equation \((2^{2m+1}-1)+(13)^n=z^2\)

1. Introduction

The class of Diophantine equations is classified in two categories, one is linear Diophantine equations and the other one is non-linear Diophantine equations. Both categories have numerous applications in solving the puzzle problems. Aggarwal et al., [1] discussed the Diophantine equation \(223^x+241^y=z^2\) for solution. Existence of solution of Diophantine equation \(181^x+199^y=z^2\) was given by Aggarwal et al., [2]. Bhatnagar and Aggarwal [3] proved that the exponential Diophantine equation \(421^p+439^q=r^2\) has no solution in whole number.

Gupta and Kumar [4] gave the solutions of exponential Diophantine equation \(n^x+(n+3m)^y=z^{2k}\). Kumar et al., [5] studied exponential Diophantine equation \(601^p+619^q=r^2\) and proved that this equation has no solution in whole number. The non-linear Diophantine equations \(61^x+67^y=z^2\) and \(67^x+73^y=z^2\) are studied by Kumar et al., [6]. They determined that the equations \(61^x+67^y=z^2\) and \(67^x+73^y=z^2\) are not solvable in non-negative integers. The Diophantine equations \(31^x+41^y=z^2\) and \(61^x+71^y=z^2\) were examined by Kumar et al., [7]. They proved that the equations \(31^x+41^y=z^2\) and \(61^x+71^y=z^2\) are not solvable in whole numbers.

Mishra et al., [8] gave the existence of solution of Diophantine equation \(211^{\alpha}+229^{\beta}=\gamma^2\) and proved that the Diophantine equation \(211^{\alpha}+229^{\beta}=\gamma^2\) has no solution in whole number. Diophantine equations help us for finding the integer solution of famous Pythagoras theorem and Pell’s equation [9,10]. Sroysang [11,12,13,14] studied the Diophantine equations \(8^x+19^y=z^2\) and \(8^x+13^y=z^2\). He determined that \(\{x=1,y=0,z=3 \}\) is the unique solution of the equations \(8^x+19^y=z^2\) and \(8^x+13^y=z^2\). Sroysang [12] studied the Diophantine equation \(31^x+32^y=z^2\) and determined that it has no positive integer solution. Sroysang [13] discussed the Diophantine equation \(3^x+5^y=z^2\).

Goel et al., [15] discussed the exponential Diophantine equation \(M_{5}^p+M_{7}^q=r^2\) and proved that this equation has no solution in whole number. Kumar et al., [16] proved that the exponential Diophantine equation \((2^{2m+1}-1)+(6^{r+1}+1)^n=w^2\) has no solution in whole number. The exponential Diophantine equation \((7^{2m} )+(6r+1)^n=z^2\) has studied by Kumar et al., [17]. Aggarwal and Sharma [18] studied the non-linear Diophantine equation \(379^x+397^y=z^2\) and proved that this equation has no solution in whole number. To determine the solution of exponential Diophantine equation \((2^{2m+1}-1)+(13)^n=z^2\), where \(m\), \(n\) are whole numbers, in whole numbers is the main objective of this paper.

2. Preliminaries

Lemma 1. For any whole number \(m\), the exponential Diophantine equation \((2^{2m+1}-1)+1=z^2\) is not solvable in whole number.

Proof. For any whole number \(m\), \(2^{2m+1}\) is an even number so \((2^{2m+1}-1)+1=z^2\) is an even number implies \(z\) is an even number. Which means

Now, for the same \(m\), \(2^{2m+1}\equiv 2(mod3)\) implies; The result of (2) denies the result of (1), hence the exponential Diophantine equation \((2^{2m+1}-1)+1=z^2\) is not solvable in whole number.

Lemma 2. For any whole number \(n\), the exponential Diophantine equation \(1+(13)^n=z^2\) is not solvable in whole number.

Proof. For any whole number \(n\), \((13)^n\) is an odd integer, so \(1+(13)^n=z^2\) is an even integer, implies \(z\) is an even integer. Which means

Now, for the same \(n\) \(13^n\equiv1(mod3)\) implies; The result of (4) denies the result of (3), hence the exponential Diophantine equation \(1+(13)^n=z^2\) is not solvable in whole number.

Theorem 1. For any whole numbers \(m\) and \(n\), the exponential Diophantine equation \((2^{2m+1}-1)+(13)^{n}=z^{2}\) is not solvable in whole number.

Proof. We have following four cases;

• 1. If \(m=0\) then the exponential Diophantine equation \((2^{2m+1}-1)+(13)^n=z^2\) becomes \(1+(13)^n=z^2\), which is not solvable in whole numbers according to Lemma 2.
• 2. If \(n=0\) then the exponential Diophantine equation \((2^{2m+1}-1)+(13)^n=z^2\) becomes \((2^{2m+1}-1)+1=z^2\), which is not solvable in whole numbers according to Lemma 1.
• 3. If \(m,n\) are natural numbers then \((2^{2m+1}-1),(13)^n\) are odd numbers, so \((2^{2m+1}-1)+(13)^n=z^2\) is an even number, implies \(z\) is an even number. which means
  \begin{equation} \label{e5} z^2\equiv0(mod3)\ \ \ \text{or}\ \ \ z^2\equiv1(mod3). \end{equation}

  (5)
  Now, \(2^{2m+1} \equiv2(mod3)\) and \((13)^n\equiv1(mod3)\) implies
  \begin{equation} \label{e6} (2^{2m+1}-1)+(13)^n=z^{2}\equiv 2(mod3). \end{equation}

  (6)
  The result of (6) denies the result of (5) hence the exponential Diophantine equation \((2^{2m+1}-1)+(13)^n=z^2\), where \(m\), \(n\) are positive integers, is not solvable in whole numbers.
• 4. If \(m,n=0\) then \((2^{2m+1}-1)+(13)^n=1+1=2=z^2\), which is impossible because \(z\) is a whole number. Hence the exponential Diophantine equation \((2^{2m+1}-1)+(13)^n=z^2\), where \(m,n=0\) is not solvable in whole numbers.

3. Conclusion

In this paper, author fruitfully studied the exponential Diophantine equation \((2^{2m+1}-1)+(13)^n=z^2\), where \(m,n\) are whole numbers, for its solution in whole number. Author determined that the exponential Diophantine equation \((2^{2m+1}-1)+(13)^n=z^2\), where \(m,n\) are whole numbers, is not solvable in whole numbers.

Conflicts of Interest

The author declares no conflict of interest.