24. N2 (COL) Consider the system
$$
\begin{aligned}
x+y & =z+u, \\
2 x y & =z u .
\end{aligned}
$$

Find the greatest value of the real constant $m$ such that $m \leq x / y$ for every positive integer solution $x, y, z, u$ of the system with $x \geq \bar{y}$.
24. We shall find the general solution to the system. Squaring both sides of the first equation and subtracting twice the second equation we obtain $(x-y)^{2}=z^{2}+u^{2}$. Thus $(z, u, x-y)$ is a Pythagorean triple. Then it is well known that there are positive integers $t, a, b$ such that $z=t\left(a^{2}-b^{2}\right)$, $u=2 t a b$ (or vice versa), and $x-y=t\left(a^{2}+b^{2}\right)$. Using that $x+y=z+u$ we come to the general solution:
$$
x=t\left(a^{2}+a b\right), \quad y=t\left(a b-b^{2}\right) ; \quad z=t\left(a^{2}-b^{2}\right), \quad u=2 t a b .
$$

Putting $a / b=k$ we obtain
$$
\frac{x}{y}=\frac{k^{2}+k}{k-1}=3+(k-1)+\frac{2}{k-1} \geq 3+2 \sqrt{2},
$$
with equality for $k-1=\sqrt{2}$. On the other hand, $k$ can be arbitrarily close to $1+\sqrt{2}$, and so $x / y$ can be arbitrarily close to $3+2 \sqrt{2}$. Hence $m=3+2 \sqrt{2}$.
Remark. There are several other techniques for solving the given system. The exact lower bound of $m$ itself can be obtained as follows: by the system $\left(\frac{x}{y}\right)^{2}-6 \frac{x}{y}+1=\left(\frac{x-u}{y}\right)^{2} \geq 0$, so $x / y \geq 3+2 \sqrt{2}$.