Problem 1549

1. Given Equation and Initial Setup:
We start with the equation $xy - z^2 = 1$ for positive integers $x, y, z$. We aim to show that there exist non-negative integers $a, b, c, d$ such that:
$$x = a^2 + b^2, \quad y = c^2 + d^2, \quad z = ac + bd$$

2. **Factorization in $\mathbb{Z}[i]$:**
Consider the factorization in the Gaussian integers $\mathbb{Z}[i]$:
$$xy = (z+i)(z-i)$$
By the unique factorization property in $\mathbb{Z}[i]$, we can write:
$$z+i = u_1 v_2, \quad z-i = u_2 v_1$$
for some non-zero $u_1, u_2, v_1, v_2 \in \mathbb{Z}[i]$.

3. **Expressing $x$ and $y$:**
Since $x$ and $y$ are real, we can write:
$$x = u_1 \overline{u_1}, \quad y = u_2 \overline{u_2}$$
where $\overline{u_1}$ and $\overline{u_2}$ are the conjugates of $u_1$ and $u_2$, respectively.

4. **Rational Numbers $q_1$ and $q_2$:**
We express $v_1$ and $v_2$ in terms of $u_1$ and $u_2$:
$$v_1 = \frac{x}{u_1} = \frac{x \overline{u_1}}{N(u_1)} = q_1 \overline{u_1}, \quad v_2 = \frac{y}{u_2} = \frac{y \overline{u_2}}{N(u_2)} = q_2 \overline{u_2}$$
where $q_1$ and $q_2$ are positive rational numbers.

5. **Equating $z+i$ and $z-i$:**
Substituting these into the equations for $z$ gives:
$$z+i = q_2 u_1 \overline{u_2}, \quad z-i = q_1 \overline{u_1} u_2$$
Since $z+i$ and $z-i$ are conjugates, the right-hand sides must also be conjugates. Given that $q_1$ and $q_2$ are real, we must have $q_1 = q_2$.

6. **Common Value $k/\ell$:**
Let the common value be $\frac{k}{\ell}$ for coprime positive integers $k$ and $\ell$. Then:
$$\ell z + \ell i = k \text{Re}(u_1 \overline{u_2}) + i k \text{Im}(u_1 \overline{u_2})$$
From the imaginary parts, we get that $k$ divides $\ell$, so $k = 1$. Also, $\ell = \text{Im}(u_1 \overline{u_2}) \mid \text{Re}(u_1 \overline{u_2})$.

7. **Writing $u_1$ and $u_2$:**
Write $u_1 = a_0 + b_0 i$ and $u_2 = c_0 + d_0 i$. Then:
$$\ell = b_0 c_0 - a_0 d_0 \mid a_0 c_0 + b_0 d_0 = z \ell$$
and:
$$x = \frac{u_1 \overline{u_1}}{\ell} = \frac{a_0^2 + b_0^2}{\ell}$$
Thus:
$$v_1 = \frac{x}{u_1} = \frac{a_0 - b_0 i}{\ell}$$
Since $v_1 \in \mathbb{Z}[i]$, it follows that $\ell \mid a_0, b_0$; similarly, $\ell \mid c_0, d_0$.

8. **Substituting $a_0, b_0, c_0, d_0$:**
Write $a_0 = \ell a$, $b_0 = \ell b$, $c_0 = \ell c$, $d_0 = \ell d$. Then:
$$\ell = \ell^2 (bc - ad) \implies \ell (bc - ad) = 1 \implies \ell = 1$$
This implies:
$$x = a^2 + b^2, \quad y = c^2 + d^2, \quad z = \text{Re}(u_1 u_2) = ac + bd$$

9. Condition Verification:
The condition $xy = z^2 + 1$ insists on:
$$(ad - bc)^2 = xy - z^2 = 1 \implies |ad - bc| = 1$$
Conversely, any $a, b, c, d$ with $|ad - bc| = 1$ give rise to a solution $x = a^2 + b^2$, $y = c^2 + d^2$, $z = ac + bd$ to $xy = z^2 + 1$.