1BE * Find all integer solutions \(m, n\) that satisfy \(m^{4}+(m+1)^{4}=n^{2}+(n+1)^{2}\).
From the original equation, we get \( m(m+1)\left(m^{2}+m+2\right)=n(n+1) \). Let \( m^{2}+m=k \), then we have \( k(k+2)=n(n+1) \).

When \( k>0 \), \( n(n+1)>0 \), so \( n>0 \) or \( n+1>0 \). If \( k \geqslant n \), then \( k(k+2)>n(n+1) \); if \( n>0, k<n \), then \( k(k+2) \leqslant(n-1)(n+1)<n(n+1) \). Both are impossible. If \( n+1<0 \), let \( n_{1}=-(n+1) \), then \( k(k+2)=n_{1}\left(n_{1}+1\right) \), which is also impossible.

When \( k<-2 \), let \( k_{1}=-(k+2) \), it then reduces to the above case.

When \( k=-1 \), then \( n^{2}+n=-1 \), which has no real solutions.

When \( k=0 \) or \(-2\), \( n(n+1)=0 \), so \( n=0 \) or \(-1\).

Therefore, \((m, n)=(0,0),(0,-1),(-1,0),(-1,-1)\).