18. Let $k$, $a$, $b$ be positive integers, and the quotients when $k$ is divided by $a^2$ and $b^2$ are $m$ and $m+116$ respectively.
(1) If $a$ and $b$ are coprime, prove that $a^2 - b^2$ is coprime with both $a^2$ and $b^2$;
(2) When $a$ and $b$ are coprime, find the value of $k$;
(3) If the greatest common divisor of $a$ and $b$ is 5, find the value of $k$.
18. (1) Let $s$ be the greatest common divisor of $a^{2}-b^{2}$ and $a^{2}$.

Then $a^{2}-b^{2}=s u, a^{2}=s v\left(s, v \in \mathbf{N}_{+}\right)$.
Thus, $a^{2}-\left(a^{2}-b^{2}\right)=b^{2}=s(v-u)$.
It is clear that $s$ is a divisor of $b^{2}$.
Since $a, b$ are coprime, $a^{2}, b^{2}$ are also coprime.
Therefore, $s=1$, meaning $a^{2}-b^{2}$ and $a^{2}$ are coprime.
Similarly, $a^{2}-b^{2}$ and $b^{2}$ are coprime.
(2) Since $k=m a^{2}=(m+116) b^{2}$, we have $m\left(a^{2}-b^{2}\right)=116 b^{2}(a>b)$.
Given that $a, b, m$ are all positive integers, $a^{2}-b^{2}$ divides $116 b^{2}$.

Since $a^{2}-b^{2}$ and $b^{2}$ are coprime, $a^{2}-b^{2}$ divides 116, i.e., $(a+b)(a-b)$ divides 116.

And $116=2^{2} \times 29, a+b$ and $a-b$ have the same parity, and $a+b>a-b>0$, so,
$$
\left\{\begin{array}{l}
a+b=29, \\
a-b=1
\end{array}\right.
$$

or $\left\{\begin{array}{l}a+b=2 \times 29, \\ a-b=2 .\end{array}\right.$
Solving these, we get $a=15, b=14$ or $a=30, b=28$.
Since $a, b$ are coprime, we have $a=15, b=14$.
Thus, $m=\frac{116 b^{2}}{a^{2}-b^{2}}=2^{4} \times 7^{2}$.
Therefore, $k=m a^{2}=2^{4} \times 7^{2} \times 15^{2}=176400$.
(3) If the greatest common divisor of $a, b$ is 5, let $a=5 a_{1}, b=$ $5 b_{1}$, then $a_{1}, b_{1}$ are coprime.
Similarly to (2), we have $m\left(a^{2}-b^{2}\right)=116 b^{2}$, i.e.,
$m\left(25 a_{1}^{2}-25 b_{1}^{2}\right)=116\left(25 b_{1}^{2}\right)$.
Thus, $m\left(a_{1}^{2}-b_{1}^{2}\right)=116 b_{1}^{2}$, and $a_{1}, b_{1}$ are coprime.
According to (2),
$$
m=2^{4} \times 7^{2}, a_{1}=15, b_{1}=14 .
$$

Therefore, $k=m a^{2}=m\left(5 a_{1}\right)^{2}=25 m a_{1}^{2}=4410000$.