Harvard-MIT Mathematics Tournament

Solution:

Answer: $-3, 2, 5$

Let $f(x) = 2 x^{2} + x - 6 = (2x - 3)(x + 2)$.

Suppose a positive integer $a$ divides both $2x - 3$ and $x + 2$. Then $a$ must also divide $2(x + 2) - (2x - 3) = 7$. Hence, $a$ can either be $1$ or $7$.

As a result, $2x - 3 = 7^{n}$ or $-7^{n}$ for some positive integer $n$, or either $x + 2$ or $2x - 3$ is $\pm 1$.

We consider the following cases:

- $(2x - 3) = 1$. Then $x = 2$, which yields $f(x) = 4$, a prime power.

- $(2x - 3) = -1$. Then $x = 1$, which yields $f(x) = -3$, not a prime power.

- $(x + 2) = 1$. Then $x = -1$, which yields $f(x) = -5$, not a prime power.

- $(x + 2) = -1$. Then $x = -3$, which yields $f(x) = 9$, a prime power.

- $(2x - 3) = 7$. Then $x = 5$, which yields $f(x) = 49$, a prime power.

- $(2x - 3) = -7$. Then $x = -2$, which yields $f(x) = 0$, not a prime power.

- $(2x - 3) = \pm 7^{n}$, for $n \geq 2$. Then, since $x + 2 = \frac{(2x - 3) + 7}{2}$, 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 $f(x) = 147$.

So $x$ can be either $-3, 2$ or $5$.