Problem 1481

To prove that at most one of the equations $ax^2 - by^2 = 1$ and $ax^2 - by^2 = -1$ has solutions in positive integers, we will use properties of quadratic forms and norms in the ring $\mathbb{Z}[\sqrt{AB}]$.

1. Assume the existence of solutions:
Suppose both equations $ax^2 - by^2 = 1$ and $ax^2 - by^2 = -1$ have solutions in positive integers. Let $(x_1, y_1)$ be a solution to $ax^2 - by^2 = 1$ and $(x_2, y_2)$ be a solution to $ax^2 - by^2 = -1$.

2. **Norms in $\mathbb{Z}[\sqrt{AB}]$:**
Consider the expressions $x_1\sqrt{a} - y_1\sqrt{b}$ and $x_2\sqrt{a} - y_2\sqrt{b}$. These belong to the ring $\mathbb{Z}[\sqrt{AB}]$ and have norms:
$$N(x_1\sqrt{a} - y_1\sqrt{b}) = (x_1\sqrt{a} - y_1\sqrt{b})(x_1\sqrt{a} + y_1\sqrt{b}) = x_1^2 a - y_1^2 b = 1$$
$$N(x_2\sqrt{a} - y_2\sqrt{b}) = (x_2\sqrt{a} - y_2\sqrt{b})(x_2\sqrt{a} + y_2\sqrt{b}) = x_2^2 a - y_2^2 b = -1$$

3. Fundamental unit and powers:
Since $x_1\sqrt{a} - y_1\sqrt{b}$ and $x_2\sqrt{a} - y_2\sqrt{b}$ have norms 1 and -1 respectively, they can be expressed as powers of the fundamental unit in $\mathbb{Z}[\sqrt{AB}]$. Specifically, we can write:
$$(x_1\sqrt{a} - y_1\sqrt{b})^2 = (a - b\sqrt{AB})^m \quad \text{for some integer } m$$
$$(x_2\sqrt{a} - y_2\sqrt{b})^2 = (a - b\sqrt{AB})^n \quad \text{for some integer } n$$

4. Contradiction in norms:
If both $ax^2 - by^2 = 1$ and $ax^2 - by^2 = -1$ have solutions, then:
$$(x_1\sqrt{a} - y_1\sqrt{b})^2 = (a - b\sqrt{AB})^m \quad \text{and} \quad (x_2\sqrt{a} - y_2\sqrt{b})^2 = (a - b\sqrt{AB})^n$$
Since the norms of these expressions are 1 and -1 respectively, it implies:
$$(a - b\sqrt{AB})^m = 1 \quad \text{and} \quad (a - b\sqrt{AB})^n = -1$$
However, this is a contradiction because the norm of a power of a unit in $\mathbb{Z}[\sqrt{AB}]$ cannot be both 1 and -1 simultaneously.

5. Conclusion:
Therefore, at most one of the equations $ax^2 - by^2 = 1$ and $ax^2 - by^2 = -1$ can have solutions in positive integers.

$\blacksquare$