Solving the Diophantine Equation 2n1^2 2^k17: A Comprehensive Analysis

The study of Diophantine equations, particularly integer solutions, forms a significant part of number theory. In this article, we focus on the Diophantine equation:

2n1^2 2^k17

where n1 and k are integers. Our objective is to determine all solutions to this equation by analyzing two key cases based on the parity of k. Let's delve into the detailed steps required to find these solutions.

Case 1: Even Values of k

When k is even, there exists a positive integer p such that:

k 2p

Substituting this into the given equation, we obtain:

2n1^2 2^{2p}17

This can be further simplified to:

2n1^2 - 2^p^2 17

Let's break this down into the following systems:

2n1 - 2^p 1

2n12^p 17

2n1 - 2^p 17

2n12^p 1

2n1 - 2^p -1

2n12^p -17

2n1 - 2^p -17

2n12^p -1

Solving the Systems

System 1:

2n1 - 2^p 1

2n12^p 17

Solving the first equation for n1, we get:

2n1 2^p 1

n1 2^{p-1} 2^{p-2}

Substituting this into the second equation:

2(2^{p-1} 2^{p-2})2^p 17

2^{p1} 2^{p} 17

By trial and error, we find:

When p 3, n1 4

Thus, the solution is:

n 4, k 6

System 2:

2n1 - 2^p 17

2n12^p 1

Again, solving the first equation for n1, we get:

2n1 2^p 17

n1 2^{p-1} 8.5

No integer solution exists since n1 must be an integer.

System 3:

2n1 - 2^p -1

2n12^p -17

Similar to the previous systems, no integer solution exists.

System 4:

2n1 - 2^p -17

2n12^p -1

Solving the first equation for n1 and substituting into the second equation, we again find no integer solution.

Case 2: Odd Values of k

When k is odd, there exists a positive integer q such that:

k 2q1

Substituting this into the original equation, we get:

2n1^2 2^{2q1}17

This simplifies to:

2n1^2 - 2^{2q1}17 4n1^2 - 2^{2q1}17 n1^2 - 2^{2q}3^2 0

This is a special case of the Diophantine equation:

x^2 - y^2 z^2 - t^2

where x n1, y n, z 2^q, and t 3. The complete parametric solution to this equation is given by:

x abcd, y ac - bd, z ac - bd, t ab - cd

For the original equation, the only integral solutions are:

q -12, n1 -5, k -12 q -4, n1 -5, k -5 q -3, n1 -5, k -3 q 2, n1 9, k 2 q 3, n1 9, k 3 q 11, n1 9, k 11

Conclusion

The complete set of integral solutions to the Diophantine equation 2n1^2 2^k17 is:

n1 -5, k -12 n1 -5, k -5 n1 -5, k -3 n1 9, k 2 n1 9, k 3 n1 9, k 4 n1 9, k 11

The case of odd k is addressed using the lemma that the Diophantine equation 2^xb^t c^z admits a solution when x ≥ 1 and t 1 when z is even, if and only if 2^x ≥ b^{50/13}. This provides an upper bound for x, which is then subjected to computer verification to find all possible solutions.