Selection Examination A

For $n=0$, the equation becomes: $m^4+m^2-m+1=0$, which is impossible, because $m^4+m^2-m+1=m^4+m(m-1)+1>0$, for all $m\in {\mathbb{Z}}_{+}$.

For $m=0$, the equation becomes: $n(n+2)=4$, impossible in the set of non-negative integers.

For $m,n\ne 0$, the equation becomes: $$n^2-2n-4(m^4+m^2-m+1)=0,(1)$$ which is of second degree with respect to $n$ and in order to have a root in integers its discriminant must be a perfect square, i.e. $$\begin{array}{rl}\Delta & =4(1+4m^4+4m^2-4m+4)=\text{perfect square}\\ & \iff 4m^4+4m^2-4m+5=\text{perfect square.}\end{array}$$ However we have $4m^4+4m^2-4m+5=(2m^2{)}^2+4m(m-1)+5>(2m^2{)}^2$, for every positive integer $m$, and also we have: $$4m^4+4m^2-4m+5=(2m^2+1{)}^2-4(m-1)\le (2m^2+1{)}^2,$$ where equality holds only for $m=1$. Hence the unique possible value for $m$ is $m=1$. Then the given equation becomes $n(n+2)=8\iff n=2$, and the pair $(m,n)=(1,2)$ is a solution.