12-35 How many integer pairs $(m, n)$ satisfy $n^{2}+(n+1)^{2}=m^{4}+(m+1)^{4}$?
(China Beijing High School Grade 1 Mathematics Competition, 1988)
［Solution］The original expression can be transformed into
$$
n^{2}+(n+1)^{2}-2 n(n+1)+2 n(n+1)
$$
$$
\begin{aligned}
= & m^{4}+(m+1)^{4}-2 m^{2}(m+1)^{2}+2 m^{2}(m+1)^{2}, \\
& 1+2 n(n+1)=\left[m^{2}-(m+1)^{2}\right]^{2}+2 m^{2}(m+1)^{2}, \\
& 1+2 n(n+1)=(2 m+1)^{2}+2 m^{2}(m+1)^{2}, \\
& 1+2 n(n+1)=4 m^{2}+4 m+1+2 m^{2}(m+1)^{2}, \\
& n(n+1)=m(m+1)[m(m+1)+2] .
\end{aligned}
$$

Let $m(m+1)=k$, then the original expression becomes
$$
n(n+1)=k(k+2),
$$

The above equation only holds when both sides are simultaneously 0, i.e., $n=0$ or $n=-1$, $k=0$ or $k=-2$, which leads to $m=0$ or $m=-1$.
Therefore, the integer pairs that satisfy the equation are only four:
$$
(m, n)=(0,0),(0,-1),(-1,0),(-1,-1)
$$