Solution
By rewriting the equation in the form
$$
m^{4}-42 m^{2}+425=n^{4}-8 n^{2}
$$
and then completing the square on each side, we obtain
$$
\left(m^{2}-21\right)^{2}=\left(n^{2}-4\right)^{2} \text {. }
$$

Taking the square root of each side, we get
$$
m^{2}-21= \pm\left(n^{2}-4\right) \text {. }
$$

Hence there are two cases to consider.
$$
m^{2}-21=n^{2}-4
$$

In this case, we have
$$
m^{2}-n^{2}=21-4,
$$
so that
$$
(m-n)(m+n)=17 .
$$

Therefore, because 17 is prime, $m-n$ and $m+n$ are equal to 1 and 17 , or -1 and -17 , in some order.

Thus in this case there are four solutions for $(m, n)$, namely $( \pm 9, \pm 8)$.
$$
m^{2}-21=-\left(n^{2}-4\right)
$$

In this case, we have
$$
m^{2}+n^{2}=21+4
$$

Hence
$$
m^{2}+n^{2}=5^{2}
$$

Now a square is non-negative, so that $-5 \leq m, n \leq 5$.
Thus in this case there are twelve solutions for $(m, n)$, namely $(0, \pm 5),( \pm 5,0),( \pm 3, \pm 4)$ and $( \pm 4, \pm 3)$.

Therefore altogether there are sixteen solutions to the given equation.
3. How many solutions are there to the equation
$$
m^{4}+8 n^{2}+425=n^{4}+42 m^{2},
$$
where $m$ and $n$ are integers?