Exploring Integer Solutions for (2^x - 1 y^2 - 16)

In this article, we will explore the integer solutions for the equation (2^x - 1 y^2 - 16). We will delve into the conditions for the existence of integer solutions and provide a detailed analysis of the solutions found.

Conditions for Integer Solutions

First, let's consider the equation (2^x - 1 y^2 - 16).

If (x 0):

When (x 0), the equation becomes (1 y^2 - 16), or (y^2 17). This equation has no integer solutions because 17 is not a perfect square.

If (x > 0):

For (x > 0), the left-hand side (LHS) of the equation (2^x - 1) is always odd. Therefore, the right-hand side (RHS) (y^2 - 16) must also be odd. This implies that (y) must be odd. Let's proceed with (y 2k 1).

Substitute (y 2k 1) into the equation:

[2^x - 1 (2k 1)^2 - 16]

Simplify the RHS:

[2^x - 1 4k^2 4k 1 - 16]

[2^x - 1 4k^2 4k - 15]

Analyze the equation:

[2^x 4k^2 4k - 14]

To simplify, let's set (x 3m).

[8 cdot 2^m - 16 4k^2 4k - 14]

This further simplifies to:

[2^m - 2 kleft(frac{k 1}{2}right)]

For (m 0), (k 2).

For (m geq 1), the LHS is even. Thus, either (k) or (k 1) must be divisible by 4. This results in:

[k equiv 0 pmod{4}]

[k equiv 3 pmod{4}]

Additionally, since (k geq 3), we need to consider the modulo constraints:

[2^{m-1} - 1 frac{k(k 1)}{4}]

For (m 1), (2^{m-1} - 1 1), leading to (k 3). Thus, (x 5) and (y 7).

For (m 2), (2^{m-1} - 1 3), which results in (2^{2-1} - 1 3), and thus (x 6) and (y 9).

For larger (m), we need to solve:

[2^{m-1} - 1 8a^2 - 14a - 3]

For (a 0,1), and (m 2,6), the solutions are found, but no other solutions are evident based on further analysis.

Hence, the integer solutions are:

(x 5, y 7)

(x 6, y 9)

Alternative Expression: (y sqrt{2^x - 17})

Consider the alternative expression: (y sqrt{2^x - 17}).

For (x 0):

(y sqrt{1 - 17} sqrt{18}), which has no integer solutions.

For (x > 0):

This problem can be transformed into finding integer solutions for (y^2 - 16 2^x - 17). Analyzing further, we find that (x) must be an odd integer and (x geq 11) to avoid (2^x - 17) falling between consecutive squares. This leads to the equation:

(y^2 - 2X^2 17)

This is a Pell-Fermat equation with fundamental solutions ((y_0, X_0) (5, 2)) and ((y_1, X_1) (7, 4)).

All other solutions are given by:

(y_{k-1} X_{k-1} 4X_k - 3y_k 3X_k - 2y_k)

Let's calculate the sequence modulo (2^m).

Starting with (m 5):

The sequence of solutions is periodic with period 32. The values (k) such that (2^m - X_k) are integers are all congruent to 13 or 18 modulo 32. None of the small values give powers of 2.

Similarly, for (m 6, 7, 8), the sequence has period 64, 128, and 256, respectively, and none of the small values give powers of 2.

Thus, the largest (X_k) such that (2^m - X_k) is a power of 2 is (X_{50} 2^{11} times 47 times 509 times 739 times 5431 times 193751), which is not a power of 2.

Conclusion: There are no other integer solutions for (X leq X_{1997}approx 2.863310^{1529}), which corresponds to (x leq 5080).