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 \( x \geq y \).
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 \(\operatorname{system}\left(\frac{x}{y}\right)^{2}-6 \frac{x}{y}+1=\left(\frac{z-u}{y}\right)^{2} \geq 0\), so \(x / y \geq 3+2 \sqrt{2}\).