Solving the Exponential Diophantine Equation 5^x y^2 1 with Techniques from Ring Theory

In this article, we explore the solutions to the exponential Diophantine equation

5x#8722; 5 y2 - 4

with x, y in mathbb{N}. The objective is to determine the integer solutions, if any, for this particular equation.

Initial Transformations

The given equation,

5x#8722;5 y2 - 4

can be factored and transformed into the following form:

5(y 1)(y-1) (y 2)(y-2)

From this, we can derive several cases based on the factorization:

Case 1: y 0, 2, or -2. These values are immediately excluded since y in mathbb{N}. Case 2: Set y_0 7, resulting in x_0 1. Case 3: Set y_1 3, resulting in a contradiction since the equation holds for only the trivial solution x 1. Exclude this as well.

Advanced Analysis with Ring Theory

To proceed further, we consider the given expression within the ring of Gaussian integers mathbb{Z}[i]. Recall that 5 (2 - i)(2 i). By the fundamental theorem of arithmetic for Gaussian integers, we can factorize both sides of the given equation as:

2-i^n (2 i)^n a-i a ai -1 i

Define g gcd(2 - i, 2 i). This means g 1, 2, 2i, or -2i. Since 2 is not a factor of the Gaussian integers in question, we conclude that g 1.

From this, we can deduce that the factors on both sides are coprime. Assuming a eq 0, n eq 0, we have two cases to consider:

Case 1: 2i^n a-i a i

This case simplifies to evaluating (2i)^2 3 4i and (2i)^3 2 - 11i equiv 2 - i pmod{10}. The periodic nature of (2i)^n modulo 10 shows that this case is impossible.

Case 2: 2i^n a-i a i

Consider the evenness of n. If n is even, then x must also be even, leading to a contradiction. Assume n 2^r m 1 where m is odd and r geq 1.

Subcase 1: r geq 2

In this case, use the binomial theorem to expand 2i^n and consider the imaginary part. Reducing modulo 2^{2r}, we find a contradiction.

Subcase 2: r 1

Reconsider the imaginary part, leading to a similar contradiction when reduced modulo 8.

Conclusion

From the exhaustive analysis using Gaussian integers and prime factorization, we conclude that the only solutions are x 1, y 2.

For those interested in the technical details, a detailed explanation can be found in a previous answer by Alon Amit.