11.4. Find all pairs $(m, n)$ such that any pair $(x, y)$ satisfying the equation $\frac{x}{y}=m$, satisfies the equation $(x+y)^{2}=n$.
Answer: $m=-1, n=0$.

Solution: Let the pair $(m, n)$ satisfy the condition of the problem.

Pairs $(x, y)$ equal to $(m, 1)$ and $(2m, 2)$ satisfy the first equation. Therefore,

$$
(m+1)^{2}=(2m+2)^{2}=n \Leftrightarrow (m+1)^{2}=4(m+1)^{2}=n \Leftrightarrow m=-1, n=0
$$

since $n=4n \Leftrightarrow n=0$.

Thus, the condition of the problem is so strong that only $m=-1, n=0$ can fit. But when $m=-1$, we have $x=-y$ and $(x+y)^{2}=0=n$, i.e., the condition of the problem is satisfied.

Criteria: The correct answer is found by trial or insufficiently justified why there are no other pairs - 3 points. 7 points for a complete solution.