Three. (25 points) Let $k$, $x$, $y$ be positive integers, and the quotients when $k$ is divided by $x^{2}$ and $y^{2}$ are $n$ and $n+148$, respectively.
(1) If $(x, y)=1$, prove: $x^{2}-y^{2}$ is coprime with $x^{2}$ and $y^{2}$;
(2) If $(x, y)=1$, find the value of $k$;
(3) If $(x, y)=4$, find the value of $k$.
(1) Let $\left(x^{2}-y^{2}, x^{2}\right)=t$. Then $x^{2}-y^{2}=t p, x^{2}=t q\left(p, q \in \mathbf{N}_{+}\right)$.
Thus, $y^{2}=t(q-p) \Rightarrow t \mid y^{2}$.
Since $(x, y)=1$, it follows that $\left(x^{2}, y^{2}\right)=1$.
Therefore, $t=1$.
Hence, $x^{2}-y^{2}$ and $x^{2}$ are coprime.
Similarly, $x^{2}-y^{2}$ and $y^{2}$ are also coprime.
(2) According to the problem, we have
$$
\begin{array}{l}
k=n x^{2}=(n+148) y^{2}(x>y) \\
\Rightarrow n\left(x^{2}-y^{2}\right)=148 y^{2} \\
\Rightarrow\left(x^{2}-y^{2}\right) \mid 148 y^{2} .
\end{array}
$$

From (1), we know $\left(x^{2}-y^{2}\right) \mid 148$, i.e., $(x+y)(x-y) \mid 148$.
Since $148=2^{2} \times 37$, and $x+y$ and $x-y$ have the same parity, and
$$
\begin{array}{l}
x+y>x-y>0 \\
\Rightarrow\left\{\begin{array} { l } 
{ x + y = 3 7 , } \\
{ x - y = 1 }
\end{array} \text { or } \left\{\begin{array}{l}
x+y=2 \times 37, \\
x-y=2
\end{array}\right.\right. \\
\Rightarrow\left\{\begin{array} { l } 
{ x = 1 9 , } \\
{ y = 1 8 }
\end{array} \text { or } \left\{\begin{array}{l}
x=38, \\
y=36
\end{array}\right.\right. \text { (discard) } \\
\Rightarrow n=\frac{148 y^{2}}{x^{2}-y^{2}}=2^{2} \times 18^{2}=2^{4} \times 3^{4} \\
\Rightarrow k=2^{4} \times 3^{4} \times 19^{2} .
\end{array}
$$
(3) If the greatest common divisor of $x$ and $y$ is 4, let $x=4 x_{1}$, $y=4 y_{1}$. Then $\left(x_{1}, y_{1}\right)=1$.
According to (2), we have $n\left(x^{2}-y^{2}\right)=148 y^{2}$.
Thus, $n\left(x_{1}^{2}-y_{1}^{2}\right)=148 y_{1}^{2}$, and $\left(x_{1}, y_{1}\right)=1$.
Therefore, $n=2^{4} \times 3^{4}, x_{1}=19, y_{1}=18$.
Hence, $k=n x^{2}=16 n x_{1}^{2}=2^{8} \times 3^{4} \times 19^{2}$.