, 2012

Suppose $3mn + 3m = k(2m^2 + n^2)$ where $k$ is a natural number. The inequality between the arithmetic and geometric means gives
$$3mn + 3m = 3m(n + 1) = k(2m^2 + n^2) \geq 2\sqrt{2}kmn,$$
hence
$$k \leq \frac{3(n + 1)}{2\sqrt{2}n} = \frac{3}{2\sqrt{2}}\left(1 + \frac{1}{n}\right) \leq \frac{3}{\sqrt{2}}.$$
We conclude $k \leq 2$.

The equation $k(2m^2 + n^2) = 3mn + 3m$ may be considered a quadratic equation for $n$:
$$kn^2 - 3mn + 2km^2 - 3m = 0. \quad (6)$$
The discriminant of this equation is equal to $D = 9m^2 - 8k^2m^2 + 12km$.

If $k = 2$, then $D = 24m - 23m^2$. Since the discriminant must be non-negative, the only solution is $m = 1$ and from the quadratic equation (6) we calculate $n = 1$.

Now let $k = 1$. The discriminant of the quadratic equation (6) must be equal to a square of a non-negative integer $t$:
$$t^2 = D = m^2 + 12m = (m + 6)^2 - 36.$$
We reorganize the above equation to get $36 = (m + 6 + t)(m + 6 - t)$. Because $t \geq 0$ and $m > 0$, we get $m + 6 + t > 0$ and hence $m + 6 - t > 0$. The numbers $m + 6 + t$ and $m + 6 - t$ obviously have the same parity. We conclude $m + 6 + t = m + 6 - t = 6$ or $m + 6 + t = 18$ and $m + 6 - t = 2$. In the first case, we get $m = 0$, which is not a natural number. In the second case, we get $m = 4$ and $t = 8$, and the equation (6) has two solutions: $n = 2$ and $n = 10$.

The solutions of the problem are the following pairs of natural numbers: $(1, 1)$, $(4, 2)$ and $(4, 10)$.