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 $$2n+3=\pm p^{\alpha },3n-13=\pm p^{\beta }$$ 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: $n\equiv 1(\mathrm{mod}5)$ whereas $p=7$ implies $n\equiv 2(\mathrm{mod}7)$. 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 $(n,k,p)=(-4,3,5),(2,2,7)$.