Example 2: With $2009^{12}$ as one of the legs, and all three sides being integers, the number of different right-angled triangles (congruent triangles are considered the same) is $\qquad$.

(2009, International Mathematics Tournament of the Cities for Young Mathematicians)
Let $a$, $2009^{12}$, and $c$ be the sides of a right triangle, with $c$ being the hypotenuse. Then,
$$
(c+a)(c-a)=2009^{24}=41^{24} \times 7^{48}.
$$

Since $c+a > c-a$ and they have the same parity, both must be odd.

Also, $2009^{24}=41^{24} \times 7^{48}$ has $25 \times 49$ different factors, which can be paired into $\frac{25 \times 49-1}{2}=612$ pairs, each pair forming a possible solution.
On the other hand, $2009^{12}$ clearly does not meet the requirements.
Therefore, there are 612 pairs of different integer-sided right triangles that satisfy the conditions.