HMMT-Feb 2011 gen, Problem 8

Find all integers $x$ such that $2 x^{2}+x-6$ is a positive integral power of a prime positive integer.

Solution: Let $f(x)=2 x^{2}+x-6=(2 x-3)(x+2)$. Suppose a positive integer $a$ divides both $2 x-3$ and $x+2$. Then $a$ must also divide $2(x+2)-(2 x-3)=7$. Hence, $a$ can either be 1 or 7. As a result, $2 x-3=7^{n}$ or $-7^{n}$ for some positive integer $n$, or either $x+2$ or $2 x-3$ is $\pm 1$. We consider the following cases: \begin{itemize}   \item $(2 x-3)=1$. Then $x=2$, which yields $f(x)=4$, a prime power.   \item $(2 x-3)=-1$. Then $x=1$, which yields $f(x)=-3$, not a prime power.   \item $(x+2)=1)$. Then $x=-1$, which yields $f(x)=-5$ not a prime power.   \item $(x+2)=-1$. Then $x=-3$, which yields $f(x)=9$, a prime power.   \item $(2 x-3)=7$. Then $x=5$, which yields $f(x)=49$, a prime power.   \item $(2 x-3)=-7$. Then $x=-2$, which yields $f(x)=0$, not a prime power.   \item $(2 x-3)= \pm 7^{n}$, for $n \geq 2$. Then, since $x+2=\frac{(2 x-3)+7}{2}$, we have that $x+2$ is divisible by 7 but not by 49. Hence $x+2= \pm 7$, yielding $x=5,-9$. The former has already been considered, while the latter yields $f(x)=147$. \end{itemize} So $x$ can be either $-3,2$ or 5 . (Note: In the official solutions packet we did not list the answer -3 . This oversight was quickly noticed on the day of the test, and only the answer $-3,2,5$ was marked as correct. $\fbox{-3,2,5}$.

Answer: -3,2,5