NZMO Round One

Solution:
The equation rearranges to be
$$p^{n} = 2x(x + 5) - 3(x - 1) = 2x^{2} + 7x + 3 = (2x + 1)(x + 3).$$
Since $x$ is a non-negative integer, both factors $(2x + 1)$ and $(x + 3)$ must be positive integers. Therefore both $(2x + 1)$ and $(x + 3)$ are both powers of $p$. Let
$$2x + 1 = p^{a}$$
$$x + 3 = p^{b}.$$
Now note that $\gcd (2x + 1, x + 3) = \gcd (x - 2, x + 3) = \gcd (5, x + 3)$ which must equal either $1$ or $5$ (because $5$ is prime). We consider each case individually:

If $\gcd (2x + 1, x + 3) = 5$ then $p = 5$ and $\min (a, b) = 1$. Thus either $2x + 1 = 5$ or $x + 3 = 5$. Either way we get $x = 2$. This leads to $p^{n} = 25$ and so $(x, p, n) = (2, 5, 2)$.

If $\gcd (2x + 1, x + 3) = 1$ then $\min (a, b) = 0$. Thus $2x + 1 = 1$ or $x + 3 = 1$. So either $x = 0$ or $x = -2$. We can't have $x < 0$ so we must have $x = 0$. This leads to $p^{n} = 3$ and so $(x, p, n) = (0, 3, 1)$.

Therefore the only solutions for $(x, p, n)$ are $(2, 5, 2)$ and $(0, 3, 1)$.