Let's start with a number-filling game. Please fill in the blanks with positive integers:
$$
\begin{array}{l}
\square^{2}+17^{2}+8^{2}=\square^{2}, \\
\square^{2}+18^{2}+12^{2}=\square^{2}, \\
\square^{2}+7^{2}+9^{2}=\square^{2} .
\end{array}
$$
How should we fill in these numbers? Let's analyze it calmly.

For example, in (1), 17 is an odd number, and 8 is an even number, so $17^{2}+8^{2}$ is an odd number, which we can write as $2k+1$; then, what number can we add to make it a perfect square? By the multiplication formula
$$
k^{2}+2k+1=(k+1)^{2}
$$

we know that filling in $k$ on the left and $k+1$ on the right will work. And $k$ can be found. Because
$$
\begin{aligned}
2k+1 & =17^{2}+8^{2}=353, \\
k & =176,
\end{aligned}
$$

that is, $176^{2}+17^{2}+8^{2}=177^{2}$.
Similarly, in (2), $18^{2}+12^{2}$ is a multiple of 4, we let $18^{2}+12^{2}=4k+4$,
$$
k=116 \text{.}
$$

Since $k^{2}+4k+4=(k+2)^{2}$,
we can fill it as
$$
116^{2}+18^{2}+12^{2}=118^{2} \text{.}
$$

For (3), we can see that $7^{2}+9^{2}$ is the sum of the squares of two odd numbers, which is a number that leaves a remainder of 2 when divided by 4, and rewriting (3) as
$$
7^{2}+9^{2}=\square^{2}-\square^{2},
$$

the difference of the squares of two integers cannot leave a remainder of 2 when divided by 4, so this problem has no solution.