Problem 1432

1. Given System of Equations:
$$x + y = z + u$$
$$2xy = zu$$
We need to find the greatest value of the real constant $m$ such that $m \le \frac{x}{y}$ for any positive integer solution $(x, y, z, u)$ of the system, with $x \ge y$.

2. Analyzing the Solutions:
We are given that all solutions $(x, y, z, u)$ such that $\gcd(z, u) = d$, $\frac{z}{d}$ is odd, and $x > y$ are in the form:
$$(x, y, z, u) = d(m(m+n), n(m-n), m^2 - n^2, 2mn)$$
where $m$ and $n$ are coprime positive integers with different parities such that $m > n$.

3. **Expressing the Ratio $\frac{x}{y}$:**
Let $q = \frac{m}{n}$. Then:
$$x = d \cdot m(m+n) = d \cdot n \cdot q(q+1)$$
$$y = d \cdot n(m-n) = d \cdot n(q-1)$$
Therefore:
$$\frac{x}{y} = \frac{d \cdot n \cdot q(q+1)}{d \cdot n(q-1)} = \frac{q(q+1)}{q-1}$$

4. Optimizing the Ratio:
We need to find the minimum value of $\frac{q(q+1)}{q-1}$ for $q > 1$. Let:
$$f(q) = \frac{q(q+1)}{q-1}$$
Simplifying $f(q)$:
$$f(q) = \frac{q^2 + q}{q-1} = q + 1 + \frac{q + 1}{q-1}$$
To find the minimum value, we can use calculus. First, find the derivative $f'(q)$:
$$f'(q) = \frac{d}{dq} \left( \frac{q^2 + q}{q-1} \right)$$
Using the quotient rule:
$$f'(q) = \frac{(2q + 1)(q-1) - (q^2 + q)(1)}{(q-1)^2}$$
Simplifying the numerator:
$$(2q + 1)(q-1) - (q^2 + q) = 2q^2 - 2q + q - 1 - q^2 - q = q^2 - 2q - 1$$
Thus:
$$f'(q) = \frac{q^2 - 2q - 1}{(q-1)^2}$$
Setting $f'(q) = 0$:
$$q^2 - 2q - 1 = 0$$
Solving this quadratic equation:
$$q = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$$
Since $q > 1$, we take:
$$q = 1 + \sqrt{2}$$

5. Calculating the Minimum Value:
Substituting $q = 1 + \sqrt{2}$ into $f(q)$:
$$f(1 + \sqrt{2}) = \frac{(1 + \sqrt{2})(2 + \sqrt{2})}{\sqrt{2}} = \frac{(1 + \sqrt{2})(2 + \sqrt{2})}{\sqrt{2}}$$
Simplifying:
$$f(1 + \sqrt{2}) = \frac{2 + 3\sqrt{2} + 2}{\sqrt{2}} = \frac{4 + 3\sqrt{2}}{\sqrt{2}} = 2\sqrt{2} + 3$$

Thus, the greatest value of $m$ such that $m \le \frac{x}{y}$ for any positive integer solution $(x, y, z, u)$ of the system, with $x \ge y$, is $3 + 2\sqrt{2}$.

The final answer is $\boxed{3 + 2\sqrt{2}}$