Slovenian Mathematical Olympiad, 2008

If $p = 2$, we have $q(x) = 2x^3 - 4x^2 - x + 2 = (x-2)(2x^2 - 1)$ and $x = 2$ is a rational root. Now, let $p$ be an odd prime. The only possible candidates for rational roots are $\pm 1, \pm p, \pm \frac{1}{2}$ and $\pm \frac{p}{2}$. Let us consider all possible cases.
Since $q(1) = 3 - 2p$ and $3 - 2p$ is odd, we have $q(1) \neq 0$. Evidently, $q(-1) = -3 \neq 0$. The expression $q(-p) = -p^2 + 2p = p(2-p)$ is non-zero because $p \neq 2$. Similarly, $q(p) = -4p^3 + p^2 = p^2(1-4p) \neq 0$ and $q(\frac{1}{2}) = \frac{3}{4} \neq 0$. It is also easy to check that $q(-\frac{1}{2}) = p - \frac{3}{4} \neq 0$ and $q(\frac{p}{2}) = \frac{-p^3-2p^2+6p}{4}$. Since $p$ is odd, the denominator of this last expression is also odd, so this expression is non-zero. The same argument shows that $q(-\frac{p}{2}) = \frac{-3p^3+2p^2+2p}{4} \neq 0$.
Thus, if $p$ is an odd prime, the polynomial $q$ has no rational roots. The only solution is $p = 2$.