8.34 Write the natural numbers on a blackboard in sequence, with the rule that when a perfect square is encountered, it is skipped and the next natural number is written instead, thus forming the sequence
$$
2,3,5,6,7,8,10,11, \cdots
$$

This sequence starts with the first number being 2, the 4th number being 6, the 8th number being 11, and so on. Following this pattern, what is the 1992nd number written on the blackboard?
(China Beijing High School Grade 1 Mathematics Competition, 1992)
[Solution] Let $f(n)$ denote the $n$-th non-square number, then we have
$$
\begin{array}{cl}
f(1)=2=1+1, & f(2)=3=2+1 \\
f(3)=5=3+2, & f(4)=6=4+2 \\
f(5)=7=5+2, & f(6)=8=6+2 \\
f(7)=10=7+3, & f(8)=11=8+3 .
\end{array}
$$

From this, we can assume
$$
f(n)=n+k
$$

where $k$ satisfies

that is,
$$
\begin{array}{c}
k^{2}<f(n)<(k+1)^{2}, \\
k^{2}<n+k<(k+1)^{2} . \\
k^{2}+1 \leqslant n+k \leqslant(k+1)^{2}-1, \\
\left(k-\frac{1}{2}\right)^{2}+\frac{3}{4}=k^{2}-k+1 \leqslant n \leqslant k^{2}+k=\left(k+\frac{1}{2}\right)^{2}-\frac{1}{4} . \\
\text { Thus } \\
k-\frac{1}{2}<\sqrt{n}<k+\frac{1}{2} .
\end{array}
$$

Thus,
$k$ is the integer closest to $\sqrt{n}$, so
$$
f(n)=n+(\text { the integer closest to } \sqrt{n} \text { }) \text {. }
$$

Since the integer closest to $\sqrt{1992}$ is 45, then
$$
f(1992)=1992+45=2037 .
$$