12.66 In the interval $1 \leqslant n \leqslant 10^{6}$, how many integers $n$ are there such that the equation $n=x^{v}$ has non-negative integer solutions $x, y$, and $x \neq n$.
(Shanghai, China High School Mathematics Competition, 1990)
[Sol] Let $N\left(x^{y}\right)$ denote the number of integers $x^{y}$. If
$$
110^{6}$, then by the principle of inclusion-exclusion, we have
$$
\begin{aligned}
N\left(x^{y}\right)= & N\left(x^{2}\right)+N\left(x^{3}\right)+N\left(x^{5}\right)+N\left(x^{7}\right)+N\left(x^{11}\right)+N\left(x^{13}\right) \\
& +N\left(x^{17}\right)+N\left(x^{19}\right)-N\left(x^{6}\right)-N\left(x^{10}\right)-N\left(x^{14}\right) \\
& -N\left(x^{15}\right)
\end{aligned}
$$

Since there are $10^{3}-1$ perfect squares greater than 1 and not exceeding $10^{6}$, we have
$$
N\left(x^{2}\right)=999 \text {. }
$$

There are $10^{2}-1$ perfect cubes greater than 1 and not exceeding $10^{6}$, i.e., $N\left(x^{3}\right)=99$. Since
$$
15^{5}=8193751$ and $y=0$.
Thus, there are 1111 integers $n$ that satisfy the condition.