Slovenija 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 $q(1)\ne 0$. Evidently, $q(-1)=-3\ne 0$. The expression $q(-p)=-p^2+2p=p(2-p)$ is non-zero because $p\ne 2$. Similarly, $q(p)=-4p^3+p^2=p^2(1-4p)\ne 0$ and $q(\frac{1}{2})=\frac{3}{4}\ne 0$. It is also easy to check that $q(-\frac{1}{2})=p-\frac{3}{4}\ne 0$ 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 $q(-\frac{p}{2})=\frac{-3p^3+2p^2+2p}{4}\ne 0$. Thus, if $p$ is an odd prime, the polynomial $q$ has no rational roots. The only solution is $p=2$.