
1. Find all pairs of positive integers $m$ and $n$ for which

$$
\sqrt{m^{2}-4}<2 \sqrt{n}-m<\sqrt{m^{2}-2} .
$$


Solution. If $m, n$ satisfy the given inequalities, then clearly

$$
m \geqslant 2 \text { and } 2 \sqrt{n}-m>0 \text {, }
$$

otherwise the square roots would not be defined or the middle term $2 \sqrt{n}-m$ would not be positive, and thus could not be greater than the nonnegative expression $\sqrt{m^{2}-4}$.

Assuming that the conditions (1) are satisfied, we can make the following equivalent manipulations of the given inequalities (note that in each of the four squarings both sides are defined and nonnegative, and equally both divisions are by the positive number $n$, so all these manipulations are correct):

$$
\begin{array}{rlrlrl}
2 \sqrt{n}-m & 0, \\
2 \sqrt{m^{2}-2} & >m,\left.\quad\right|^{2} \\
4\left(m^{2}-2\right) & >m^{2}, \\
3 m^{2} & >8 .
\end{aligned}
$$

The last inequality holds, however, for any $m \geqslant 2$.

Answer. The sought pairs are precisely those of the form $(m, n)=\left(m, m^{2}-2\right)$, where $m \geqslant 2$ is any positive integer.
