Question 186: Fill the integers from $1 \sim 10000$ into a $100 \times 100$ table in natural order, with each row from left to right being: the first row 1 to 100, the second row $101 \sim 200, \cdots$, the hundredth row $9901 \sim 10000$. If the row and column containing number A have no perfect squares, then A is called a free number. How many free numbers are there in the table?
Question 186, Solution: We refer to rows (columns) that do not contain perfect squares as blank rows (columns). We will prove separately that the table has exactly 25 blank rows and 78 blank columns.
(1) When $(k+1)^{2}-k^{2}=2 k+1 \leq 100$, i.e., $k \leq 49$, $k^{2}$ and $(k+1)^{2}$ are in the same row or adjacent rows, so there are no blank rows between $1^{2}$ and $50^{2}$. Since $50^{2}=2500$ is in the 25th row, the first 25 rows have no blank rows.

In the last 75 rows, there are $100-50$ perfect squares. Since for $k \geq 50$, $(k+1)^{2}-k^{2}>100$, any two perfect squares are in different rows, so the last 50 perfect squares occupy exactly 50 rows. Therefore, there are $100-25-50=$ 25 blank rows.
(2) The numbers in the $j$-th column are congruent to $j$ modulo 100, so there is a perfect square in the column if and only if there exists $x \in$ $\{1,2, \ldots, 100\}$ such that $j \equiv x^{2}(\bmod 100) \Leftrightarrow\left\{\begin{array}{l}j \equiv x^{2}(\bmod 4) \\ j \equiv x^{2}(\bmod 25)\end{array} \circ\right.$ $x^{2} \equiv 0, 1, 4, 6, 9, 11, 14, 16, 19, 21, 24(\bmod 4)$, according to the Chinese Remainder Theorem, there are exactly $2 \times 11=22$ numbers in $\{1,2, \ldots, 100\}$ that are congruent to a square number, meaning that there are exactly 22 columns containing square numbers, and thus exactly 78 blank columns.
Free numbers are at the intersection of blank rows and blank columns, so there are $25 \times 78=1950$ free numbers.
Note: For an odd prime $p$, the quadratic residues modulo $p^{2}$ (including 0) are exactly $1+\frac{p^{2}-1}{2}-\left[\frac{\frac{p^{2}-1}{2}}{p}\right]=1+\frac{p^{2}-1}{2}-$ $\frac{p-1}{2}=1+\frac{p(p-1)}{2}$. This is because, for any integer $x, x^{2} \equiv\left(p^{2}-x\right)^{2}\left(\bmod p^{2}\right)$; if $1 \leq i<j \leq \frac{p^{2}-1}{2}$, and $(i, p)=(j, p)=1$, then $i^{2} \neq j^{2}(\bmod p)$; and when $p \mid i$, $i^{2} \equiv 0\left(\bmod p^{2}\right)$. Also, among 1, 2, ..., $\frac{p^{2}-1}{2}$, there are exactly $\left[\frac{p^{2}-1}{2}\right]$ numbers that are multiples of $p$. Therefore, the quadratic residues modulo $p^{2}$ (including 0) are exactly $1+$ $\frac{p^{2}-1}{2}-\left[\frac{\frac{p^{2}-1}{2}}{p}\right]=1+\frac{p^{2}-1}{2}-\frac{p-1}{2}=1+\frac{p(p-1)}{2}$.