XVII OBM

First we prove that $P(x)=x^5-x^4-4x^3+4x^2+2$ is irreducible in $\mathbb{Z}$. Its possible rational roots are $\pm 1,\pm 2$. Since $P(x)=x^2(x-1)(x-2)(x+2)+2$ it is clear that $\pm 2$ and $1$ are not roots and $P(-1)\ne 0$ too. So if $P(x)$ is reducible then it must be written as a product of two nonlinear polynomials. Looking $P(x)$ modulo $2$ we have $P(x)=x^4(x-1)$. Since there is unique factoring in polynomials in $\mathbb{Z}/2\mathbb{Z}$ then if $P(x)$ is reducible then $P(x)=(x^k+2P_1(x))(x^l(x-1)+2P_2(x))$, $P_1(x)$ and $P_2(x)$ with integer coefficients and $k,l>0$. But then $P(0)=4P_1(x)P_2(x)⟺P_1(x)P_2(x)=\frac{1}{2}$, which is not possible. So $P(x)$ is irreducible.

Let $\alpha$ be a root of $P(x)$. Then $P(x)$ is the minimal polynomial of $\alpha$, and if ${\alpha }^n=r$, $r$ rational, then $P(x)$ divides $Q(x)=x^n-r$. But looking modulo $2$ again we have $x^5\equiv x^4(\mathrm{mod}P(x))$, so $x^n\equiv x^4(\mathrm{mod}P(x))$ and so $\alpha$ is a root of $x^4-r$. This is a contradiction, since the minimal polynomial of $\alpha$ has degree $5$.