The first equation squared minus four times the second gives

$$
\left(\frac{x}{y}\right)^{2}-6\left(\frac{x}{y}\right)+1=\left(\frac{z-t}{y}\right)^{2}
$$

The left-hand side is a quadratic in $x / y$, which is positive given the right-hand side. Note that $x / y > 1$ according to the problem statement. The function $g: m \mapsto m^{2}-6 m+1$ is negative for $1 \leqslant m \leqslant r_{1}$ and positive for $m \geqslant r_{1}$, where $r_{1}=3+2 \sqrt{2}$ is the larger root of the quadratic. Therefore, $x / y \geqslant 3+2 \sqrt{2}$. It remains to show that there exist solutions such that $\frac{z-t}{y}$ is arbitrarily small, which will imply that $x / y$ is arbitrarily close to $r_{1}$. Note that this is not due to the continuity of $g$, but to the fact that for any $\delta>0$, there exists $\varepsilon>0$ such that if $w>r_{1}+\delta, g(w)>\varepsilon$ (exercise: prove this) and we will have shown that $m=3+2 \sqrt{2}$.

We revisit the statement: the first equation squared minus two times the second gives

$$
(x-y)^{2}=z^{2}+t^{2}
$$

Without loss of generality, we can assume $\operatorname{gcd}(x-y, z, t)=1$, and that $t$ is even. A useful theorem on Pythagorean triples (proof left to the reader) states that there exist $a>b$ coprime such that

$$
x-y=a^{2}-b^{2}, t=2 a b, z=a^{2}-b^{2}
$$

up to swapping $z$ and $t$. With $x+y=z+t$, we get $y=b(a-b)$. Can we have $z-t=1$? This amounts to solving $a^{2}-b^{2}-2 a b=1$, or $(a-b)^{2}-2 b^{2}=1$. The Pell equation $M^{2}-2 N^{2}=1$ has infinitely many solutions, so for $b=N$ and $a=N+M, z-t=1$. Moreover, $y=M N$ can be arbitrarily large, in which case $\frac{z-t}{y}=\frac{1}{M N}$ is arbitrarily small, which concludes the proof.
Let $(x, y, z, t) \in\left(\mathbb{N}^{*}\right)^{4}$ be a quadruplet satisfying $x+y=z+t$ and $2 x y=z t$, and such that $x \geqslant y$. Find the largest value of $m$ such that $m \leqslant x / y$ for sure.