Croatia competition problems

We are given the equation
$$n^2 - 6n = m^2 + m - 10.$$
Let's rewrite it as
$$n^2 - 6n - m^2 - m + 10 = 0.$$
Group terms:
$$n^2 - m^2 - 6n - m + 10 = 0.$$
Recall that $n^2 - m^2 = (n - m)(n + m)$, so
$$(n - m)(n + m) - 6n - m + 10 = 0.$$
Let us solve for $n$ in terms of $m$.

Alternatively, move all terms to one side:
$$n^2 - 6n - m^2 - m + 10 = 0$$
$$n^2 - 6n = m^2 + m - 10$$
This is a quadratic in $n$:
$$n^2 - 6n - (m^2 + m - 10) = 0$$
The discriminant must be a perfect square for integer solutions:
$$\Delta = 36 + 4(m^2 + m - 10) = 36 + 4m^2 + 4m - 40 = 4m^2 + 4m - 4 = 4(m^2 + m - 1)$$
So $4(m^2 + m - 1)$ must be a perfect square, say $k^2$ for some integer $k$:
$$4(m^2 + m - 1) = k^2$$
So
$$k^2 - 4m^2 - 4m + 4 = 0$$
$$k^2 - 4m^2 - 4m + 4 = 0$$
$$k^2 = 4m^2 + 4m - 4$$
$$k^2 = 4(m^2 + m - 1)$$
So $m^2 + m - 1$ must be a perfect square times $1/4$.
But $k$ must be even, so let $k = 2t$ for some integer $t$:
$$(2t)^2 = 4(m^2 + m - 1)$$
$$4t^2 = 4(m^2 + m - 1)$$
$$t^2 = m^2 + m - 1$$
So $m^2 + m - 1 = t^2$ for some integer $t$.

Now, rearrange:
$$m^2 + m - 1 - t^2 = 0$$
$$m^2 + m - t^2 = 1$$
This is a quadratic in $m$:
$$m^2 + m - (t^2 + 1) = 0$$
The discriminant must be a perfect square:
$$\Delta = 1^2 - 4 \cdot 1 \cdot (-t^2 - 1) = 1 + 4t^2 + 4 = 4t^2 + 5$$
So $4t^2 + 5$ must be a perfect square, say $s^2$:
$$s^2 = 4t^2 + 5$$
So $s^2 - 4t^2 = 5$

This is a Pell-type equation.

Let us solve $s^2 - 4t^2 = 5$ in integers.

Try small integer values for $t$:
- $t = 0$: $s^2 = 5$ (no integer solution)
- $t = 1$: $s^2 = 4 + 5 = 9$, $s = 3$
- $t = -1$: $s^2 = 4 + 5 = 9$, $s = 3$
- $t = 2$: $s^2 = 16 + 5 = 21$ (no integer solution)
- $t = -2$: $s^2 = 16 + 5 = 21$ (no integer solution)
- $t = 3$: $s^2 = 36 + 5 = 41$ (no integer solution)
- $t = -3$: $s^2 = 36 + 5 = 41$ (no integer solution)
- $t = 5$: $s^2 = 100 + 5 = 105$ (no integer solution)

So only $t = 1$ and $t = -1$ work, with $s = 3$.

Now, recall $m^2 + m - t^2 = 1$
For $t = 1$:
$$m^2 + m - 1 = 1$$
$$m^2 + m = 2$$
$$m^2 + m - 2 = 0$$
$$m = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2}$$
So $m = 1$ or $m = -2$

For $t = -1$:
Same as above, $m = 1$ or $m = -2$

Now, recall $n$ is given by the quadratic:
$$n^2 - 6n = m^2 + m - 10$$
For $m = 1$:
$$n^2 - 6n = 1 + 1 - 10 = -8$$
$$n^2 - 6n + 8 = 0$$
$$n = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm 2}{2} = 4, 2$$
For $m = -2$:
$$n^2 - 6n = 4 - 2 - 10 = -8$$
So same as above, $n = 4, 2$

Therefore, the integer solutions are:
$$(m, n) = (1, 2), (1, 4), (-2, 2), (-2, 4)$$

Thus, all integer pairs $(m, n)$ such that $n^2 - 6n = m^2 + m - 10$ are:
$$(1, 2), (1, 4), (-2, 2), (-2, 4)$$