Team Selection Test for JBMO, 2023

Answer: $(n, k, p) = (-4, 3, 5), (-2, 1, 19), (-1, 4, 2), (2, 2, 7), (4, 1, 11)$.
Writing
$$p^k = |6n^2 - 17n - 39| = |(2n + 3)(3n - 13)|,$$
we get
$$2n + 3 = \pm p^{\alpha}, \quad 3n - 13 = \pm p^{\beta}$$
for some non-negative integers $\alpha, \beta$. Examining the cases $\alpha = 0$ or $\beta = 0$, we find the possibilities $n = -1, n = -2$ and $n = 4$, which gives the solutions $(n, k, p) = (-2, 1, 19), (-1, 4, 2), (4, 1, 11)$.
If $\alpha, \beta \ge 1$, we have $p|(2n+3)$ and $p|(3n-13)$, hence we get $p|35$,
so $p \in \{5,7\}$. Note that $p^{\alpha-\beta} = \pm \frac{2n+3}{3n-13}$. However, if $n \ge 17$ or $n \ge -5$, it is clear that $\frac{1}{5} < \frac{2n+3}{3n-13} < 1$, which means that $\alpha-\beta$ is not an integer as $p \in \{5,7\}$, a contradiction.
As a result, we need to examine the cases $-4 \le n \le 16$. We can eliminate some of them directly as follows. Note that if $p=5$, then we have: $n \equiv 1 \pmod 5$ whereas $p=7$ implies $n \equiv 2 \pmod 7$. Moreover, $3n-13$ is odd, so $n$ must be even. Then, we only need to check the values $n \in \{-4,2,6,16\}$, which gives the solutions $(n, k, p) = (-4, 3, 5), (2, 2, 7)$.