23. How many ordered pairs of integers $(m, n)$ where $0<m<n<2008$ satisfy the equation $2008^{2}+m^{2}=2007^{2}+n^{2} ?$
23. Ans: 3 .
Since $2008^{2}-2007^{2}=n^{2}-m^{2}$, we have $4015=(n+m)(n-m)$, i.e. $5 \cdot 11 \cdot 73=(n+m)(n-m)$. There're four possibilities: $n+m=4015, n-m=1 ; n+m=803, n-m=5 ; n+m=$ $365, n-m=11$ and $n+m=73, n-m=55$. But the first one $n=2008, m=2007$ is ruled out by assumption, the remaining pairs are $n=404, m=399 ; n=188, m=177$ and $n=64, m=9$.