12 Find all integers $m, n$ such that $m^{4}+(m+1)^{4}=n^{2}+(n+1)^{2}$.
12. After expanding and rearranging, we get
$$m^{4}+2 m^{3}+3 m^{2}+2 m=n^{2}+n$$

Adding 1 to both sides and completing the square, we get
$$\left(m^{2}+m+1\right)^{2}=n^{2}+n+1$$

When $n>0$,
$$n^{2}<n^{2}+n+1<(n+1)^{2}$$

When $n<-1$, $\quad(n+1)^{2}<n^{2}+n+1<n^{2}$,
the number $n^{2}+n+1$ is never a perfect square. Therefore, for (1) to hold, we need $n=-1$ or 0, which leads to
$$(m, n)=(-1,0),(0,0),(-1,-1),(0,-1)$$