Solving the Diophantine Equation x^2 - yx^2 - 2x - y^2 0: A Thorough Analysis

The diophantine equation x2 - yx2 - 2x - y2 0 presents a unique challenge in the realm of number theory. This equation combines elements of quadratic forms and integer solutions, making it an intriguing problem to explore. This article delves into the solution approach, highlighting the underlying methodology and solutions.

Methodology

The first step is to treat the given equation as a quadratic in x. This involves rearranging the terms to express the equation in a standard quadratic form:

$$ x^2 - yx^2 - 2x - y^2 0 $$

This can be simplified to:

$$ x^2 (1 - y) - 2x - y^2 0 $$

Here, the coefficient of x2 is (1 - y). This factor is crucial as it dictates whether the quadratic equation is solvable in integers.

Solving the Equation

The next step involves ensuring that the discriminant of this quadratic equation is a perfect square. The discriminant ((Delta)) of a quadratic equation (ax^2 bx c 0) is given by (Delta b^2 - 4ac).

For the equation (x^2 (1 - y) - 2x - y^2 0), the discriminant is:

$$ Delta (-2)^2 - 4(1 - y)(-y^2) 4 4y^3 - 4y^2 $$

This simplifies to:

$$ Delta 4 4y^3 - 4y^2 $$

For (Delta) to be a perfect square, let's denote it as a square of some integer (k), i.e., (Delta k^2). This gives us the equation:

$$ 4 4y^3 - 4y^2 k^2 $$

By simplifying further, we get:

$$ 4(y^3 - y^2 1) k^2 $$

For (4(y^3 - y^2 1)) to be a perfect square, (y^3 - y^2 1) must take specific values that make the entire expression a perfect square. This is a non-trivial condition, and solving it directly might be complex. Therefore, we can consider a simpler approach starting from the quadratic expression:

$$ y - 1 frac{2x - 1}{x^2 - 1} $$

This equation implies that for (y) to be an integer, the numerator and denominator must be such that the fraction simplifies to an integer. This can be analyzed by asserting that the numerator (2x - 1) and the denominator (x^2 - 1) must be compatible.

Detailed Analysis and Solutions

Let's rewrite the equation in a simpler form:

$$ y - 1 frac{2x - 1}{x^2 - 1} $$

This can be further simplified to:

$$ y 1 frac{2x - 1}{x^2 - 1} $$

For (y) to be an integer, the fraction (frac{2x - 1}{x^2 - 1}) must be an integer. Since (x^2 - 1 (x - 1)(x 1)), the possible values for (x) need to be such that (2x - 1) is divisible by either (x - 1) or (x 1).

Let's analyze the cases:

Case 1: (x 0)

Substituting (x 0), we get:

$$ y 1 frac{2(0) - 1}{0^2 - 1} 1 - 1 0 $$

This gives the solution ((x, y) (0, 0)).

Case 2: (x 2)

Substituting (x 2), we get:

$$ y 1 frac{2(2) - 1}{2^2 - 1} 1 frac{3}{3} 2 $$

This gives the solution ((x, y) (2, 2)).

Case 3: (x -2)

Substituting (x -2), we get:

$$ y 1 frac{2(-2) - 1}{(-2)^2 - 1} 1 - frac{5}{3} $$

This does not yield an integer value for (y).

Conclusion

Upon examining the cases, we find that the only integer solutions to the equation (x^2 - yx^2 - 2x - y^2 0) are:

$$ (x, y) (0, 0), (2, 2) $$

These solutions are derived from the condition that (frac{2x - 1}{x^2 - 1}) must be an integer.