3.501 January ** Proof: For each integer $s>1$, there exists a positive integer $m_{s}$, such that when $n \geqslant m_{s}$, there is at least one number between $n$ and $2n$ that is the $s$-th power of some integer. Find the smallest $m_{x}$ for $s=2$ and $s=3$.
Let $m_{s}=\left[\frac{1}{(\sqrt[3]{2}-1)^{s}}\right]+1$, for integers $n \geqslant m_{s}$, we have $n>\frac{1}{(\sqrt[3]{2}-1)^{s}}$, which means $\sqrt[3]{2 n}-$ $\sqrt[3]{n}=\sqrt[3]{n}(\sqrt[3]{2}-1)>1$. Therefore, there exists a positive integer $k$, such that $\sqrt[n]{n}<k<\sqrt[0]{2 n}$. That is, $n<k^{s}<2 n$. There is a square number between 5 and 10, which is $3^{2}$, while there is no square number between 4 and 8, so the smallest $m_{2}$ is 5. Similarly, the smallest $m_{3}$ is 33.