Three, (25 points) Find all integer pairs $(x, y)$ such that
$$
\begin{array}{l}
x^{3}-x^{2} y+x y^{2}-y^{3} \\
=4 x^{2}-4 x y+4 y^{2}+47 .
\end{array}
$$
$$
m-2= \pm 1, \pm 5, \pm 11, \pm 55 .
$$

Thus, $m=3,1,7,-3,13,-9,57,-53$.
(1) When $m=3$, from equation (2) we get $n=28$.

Solving simultaneously, we get $(x, y)=(7,4),(-4,-7)$.
(2) When $m=7$, from equation (2) we get $n=-10$.

Solving simultaneously, we get $(x, y)=(5,-2),(2,-5)$.
(3) When $m=1,-3,13,-9,57,-53$, from equation (2) we have $m^{2}+4 n<0$, so the system of equations has no real roots.
In summary, all the integer pairs are
$$
(x, y)=(7,4),(-4,-7),(5,-2),(2,-5) \text {. }
$$

Three, let $m=x-y, n=x y$. Substituting into the given equation, we get
$$
m\left(m^{2}+3 n\right)-n m=4 m^{2}+4 n+47,
$$

which simplifies to $2(m-2) n=-m^{3}+4 m^{2}+47$.
When $m=2$, equation (1) does not hold.
Therefore, $m \neq 2$, i.e., $m-2 \neq 0$.
Dividing both sides of equation (1) by $m-2$, we get
$$
2 n=-m^{2}+2 m+4+\frac{55}{m-2} \text {. }
$$

Since $x$ and $y$ are integers, $m$ and $n$ are also integers.
From equation (2), $\frac{55}{m-2}$ must be an integer, so

we know