Belorusija 2012

Answer: $(n;m)=(4;2)$. By condition, $$n^2+n+1=(m^2+n-3)(m^2-n+5)=m^4+m^2+8m-15.$$ Consider the obtained equation $$n^2+n-(m^4+m^2+8m-16)=0(1)$$ as a quadratic equation with respect to $n$. It has positive integer roots only if the determinant $D=4m^4+4m^2+32m-63$ of this equation is a perfect square of some integer number. But $$D=4m^4+4m^2+32m-63=(2m^2+2{)}^2-4(m-4{)}^2-3<(2m^2+2{)}^2$$ for any natural number $m$, and $$D=4m^4+4m^2+32m-63=(2m^2+1{)}^2+32(m-2)>(2m^2+1{)}^2$$ for any natural number $m>2$. Therefore, (1) has the natural roots only if $m=1$ or $m=2$. If $m=1$, then $n^2+n+6=0$, so either $n=-2$ or $n\in \mathbb{Z}$. If $m=2$, then $n^2+n-20=0$, so either $n=-5$ or $n=4$. Thus, (4; 2) is a unique pair of positive integers satisfying the problem condition.