2. find a pair of $(m, n)$ odd natural numbers with $m, n>2009$ and

$$
m\left|n^{2}+8, \quad n\right| m^{2}+8
$$

## Solution
Since $m$ and $n$ are odd, it immediately follows that $(m, n)=\left(m, m^{2}+8\right)=\left(n, n^{2}+8\right)=1$. Thus, a pair $(m, n)$ fulfills the condition of the task if $m n$ is a divisor
of $\left(m^{2}+8\right)\left(n^{2}+8\right)$. Since $m$ and $n$ are odd, this is also the case if

$$
\frac{m^{2}+n^{2}+8}{m n}=k
$$

is a natural number. Due to the symmetry of the problem, we can restrict ourselves to the pairs $(m, n)$ of odd natural numbers with $m \geq n$. We call such pairs good if even $m>n$ holds. It is immediately clear that $(1,1)$ is the only pair of solutions that is not good. In the following we fix $k$ and summarize (1) with fixed $n$ as a quadratic equation $m^{2}-k n \cdot m+\left(n^{2}+8\right)=0$ in $m$. This equation has a second solution $m^{\prime}$ and according to Vieta's theorem, the following equations apply

$$
\begin{aligned}
m+m^{\prime} & =k n \\
m m^{\prime} & =n^{2}+8
\end{aligned}
$$

From the first it follows that $m^{\prime}$ is also whole, and from the second it follows that $m^{\prime}>0$ and that $m^{\prime}$ is odd. Furthermore, if $n \geq 4$, then $m, m^{\prime}$ cannot both be greater than $n$, because if the contradiction

$$
n^{2}+8=m m^{\prime} \geq(n+1)^{2}=n^{2}+2 n+1 \geq n^{2}+9
$$

to the second equation. We therefore obtain a smaller admissible pair $\downarrow(m, n)=(n, k n-m)$ for each admissible pair $(m, n)$ with $n \geq 4$. Inversely, for each pair of solutions ( $m, n$ ) with $m \geq n$ we obtain an admissible (due to $k \geq 3$ ) pair of solutions $\uparrow(m, n)=$ $(k m-n, m)$. It is therefore sufficient to find the pairs of solutions with $m \geq n$ and $n \leq 3$. A brief analysis of these cases finally shows that all solutions arise from the two minimal solutions $(1,1)$ at $k=10$ and $(3,1)$ at $k=6$ by iterated application of $\uparrow$. For the first few iterations, the following pairs are obtained

$$
(1,1), \quad(9,1), \quad(89,9), \quad(881,89), \quad(8721,881), \ldots
$$

and

$$
(3,1), \quad(17,3), \quad(99,17), \quad(567,99), \quad(3303,567), \ldots
$$

The pairs we are looking for are therefore the first four of each of these two families (as well as their interchanges).