$ z,u$ is the root of the equation: $ t^2 - (x + y)t + 2xy$ $ \Delta = (x + y)^2 - 8xy = x^2 + y^2 - 6xy = m^2$ $ \Rightarrow (\frac {x}{y})^2 - 6\frac {x}{y} + 1 = (\frac {m}{y})^2$ $ a = \frac {x}{y},b = \frac {m}{y}$ So we has : $ a^2 - 6a + 1 = b^2$ $ \Leftrightarrow (a - 3)^2 - b^2 = 8$ $ (a - 3 - b)(a - 3 + b) = 8$ It imply that the the equation has rational solution for all $ b$ less enough. Now Let $ b\to0$ then $ a\to 3 + 2\sqrt {2}$ So the max $ m$ is $ m = 3 + 2\sqrt {2}$

A little more analysis shows that all solutions $(x,y,z,u)$ such that $\gcd(z,u)=d$, that $\frac{z}{d}$ is odd, and that $x>y$ are in the form $d\left(m(m+n),n(m-n),m^2-n^2,2mn\right)$, where $m$ and $n$ are coprime positive integers with different parities such that $m>n$. Thus, if $q$ is the ratio $\frac{m}{n}$, then $\frac{x}{y}=\frac{q(q+1)}{q-1}$. As the rationals is dense in the reals, we may seek to optimize this ratio taking $q$ to be an arbitrary real greater than $1$. Calculus or AM-GM yields the best lower bound of $3+2\sqrt2$.

Это задача Герона Александрийского. Система: $X+Y=Z+D$ $XY=qZD$ Решения можно записать: $X=qs^2-qps$ $Y=qps-p^2$ $Z=ps-p^2$ $D=qs^2-ps$ Или ещё: $X=qp^2+qps$ $Y=qps+(q-1)s^2$ $Z=qp^2+(2q-1)ps+(q-1)s^2$ $D=ps$ Где $q$ любое целое число. Видно, что система всегда имеет решения. Числа же $p,s$ задаются нами и могут быть любыми.