1. Let $A$ be a set of integers. If for any $x, y \in A$ (allowing $x, y$ to be the same), and for every integer $k$, we have $x^{2}+k x y+y^{2} \in A$, then the set $A$ is called "admissible". Find all non-zero integer pairs $(m, n)$ such that the only admissible set containing $m, n$ is the set of all integers.
1. The integer pair $(m, n)$ satisfies the condition if and only if $(m, n)=1$.
Assume $(m, n)=d>1$. Then the set
$$
A=\{\cdots,-2 d,-d, 0, d, 2 d, \cdots\}
$$

is allowed, because if $d$ divides $x, y$, then for each integer $k$, we have $d \mid\left(x^{2}+k x y+y^{2}\right)$.
Clearly, $m, n \in A, A \neq \mathbf{Z}$.
Suppose $(m, n)=1$, and $A$ is an allowed set containing $m, n$. Next, we prove:
(1) For each $x \in A$ and each integer $k$, we have $k x^{2} \in A$;
(2) For all $x, y \in A$, we have $(x+y)^{2} \in A$.
In fact, for (1) and (2), it suffices to set $y=x, k=2$ in the condition.
Since $(m, n)=1$, we have $\left(m^{2}, n^{2}\right)=1$.
By Bézout's theorem, there exist integers $a, b$ such that
$$
a m^{2}+b n^{2}=1 \text {. }
$$

By (1), we know $a m^{2} \in A, b n^{2} \in A$.
By (2), we know $1=\left(a m^{2}+b n^{2}\right)^{2} \in A$.
By (1) and $1 \in A$, we know that for each integer $k$, we have $k \in A$. Therefore, $A=\mathbf{Z}$.