10. (The 2nd Northeast China Three Provinces Mathematics Invitational Competition) Find all positive integers $m, n$, such that $(m+n)^{m}=n^{n}+1413$.
10. Since $1413+n^{m}=(m+n)^{m} \geqslant m^{m}+n^{m}$, it follows that $m^{m} \leqslant 1413$, hence $m \leqslant 4$. When $m$ is even, then $(m+n)^{m}$

has the same parity as $n$ while $1413+n^{m}$ has the opposite parity of $n$, which is impossible, so $m$ must be odd. When $m=1$, $(m+n)^{m}=1+n$ cannot equal $n^{m}+1413=n+1413$. When $m=3$, from $(n+3)^{3}=n^{3}+1413$ we get $n=11, n=-14$ (discard). Therefore, the solution is $(m, n)=(3,11)$.