Three, (25 points) If the sum, difference, product, and quotient of two unequal natural numbers add up to a perfect square, then such two numbers are called a "wise pair" (for example, $(8,2)$ is a wise pair, since $\left.(8+2)+(8-2)+8 \times 2+\frac{8}{2}=36=6^{2}\right)$.

If both of these natural numbers do not exceed 100, how many such wise pairs are there?
Three, let $(a, b)$ be a wise array, and without loss of generality, assume $a > b$.
By definition, we have
$$
(a+b)+(a-b)+a b+\frac{a}{b}=m^{2}
$$
$(a, b, m$ are natural numbers), which simplifies to
$$
2 a+a b+\frac{a}{b}=m^{2}.
$$

Since $2 a, a b, m^{2}$ are all natural numbers, $\frac{a}{b}$ must also be a natural number.
Without loss of generality, let $\frac{a}{b}=k$ (where $k$ is a natural number and $k \neq 1$). Then $a=k b$.
Substituting into equation (1) and simplifying, we get $m^{2}=(b+1)^{2} k$.
Thus, $(b+1)^{2} \mid m^{2} \Rightarrow (b+1) \mid m$.
Therefore, $k=\left(\frac{m}{b+1}\right)^{2} \geqslant 4$.
Also, $b < a \leqslant 100$, so
$1 \leqslant b \leqslant 25$, and $a=k b, 1 < k \leqslant \frac{100}{b}$.
Thus, when $b=1$, we have $k=2^{2}, 3^{2}, \cdots, 10^{2}$, so there are 9 pairs $(a, b)$ that satisfy the condition.
When $b=2$, we have $k=2^{2}, 3^{2}, \cdots, 7^{2}$, so there are 6 pairs $(a, b)$ that satisfy the condition.
Similarly, when $b=3,4$, there are 4 pairs $(a, b)$ each; when $b=5,6$, there are 3 pairs $(a, b)$ each; when $b=7,8, \cdots, 11$, there are 2 pairs $(a, b)$ each; when $b=12,13, \cdots, 25$, there is 1 pair $(a, b)$ each.
Therefore, there are a total of 53 wise arrays.
(Li Yaowen, No. 18 Middle School of Benzhuang City, Shandong Province, 277200)