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 $m, n \neq 0$, the equation becomes:
$$n^2 - 2n - 4(m^4 + m^2 - m + 1) = 0, \quad (1)$$
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.
$$\begin{aligned} \Delta &= 4(1 + 4m^4 + 4m^2 - 4m + 4) = \text{perfect square} \\ &\Leftrightarrow 4m^4 + 4m^2 - 4m + 5 = \text{perfect square.} \end{aligned}$$
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:
$$4m^4 + 4m^2 - 4m + 5 = (2m^2 + 1)^2 - 4(m-1) \leq (2m^2 + 1)^2,$$
where equality holds only for $m = 1$. Hence the unique possible value for $m$ is $m = 1$. Then the given equation becomes $n(n+2) = 8 \Leftrightarrow n = 2$, and the pair $(m, n) = (1, 2)$ is a solution.