Let $p(x)$ be polynomial over $GF(2)$ with $\operatorname{deg}(p(x))=8$, show that $p(x)\nmid x^{31}-1$
I know that $x^{2^5}-x=x^{32}-x=x(x^{31}-1)$ is a product of all monic irreducible polynomials of degree dividing 5 (actually the minimal polynomials over the field $GF(2)$). Meaning that over $GF(2)$ it is a product of only the irreducible polynomial with degree 5 and 1 when only $x$ and $x-1$ are the polynomials with degree 1. Now, it means the $x^{31}-1$ is a product of the irreducible polynomial with degree 5 and $x-1$. now assume to the contrary that $p(x)|x^{31}-1$. It means that p(x) is the product of some of the irreducible polynomials with degree 5 and $x-1$ (this is the part I'm not sure about), but it can't be since not any combination of the irreducible polynomials will give polynomial with $\deg(p(x))=8$.
As I mentioned I'm not absolutely sure about my conclusion. I know there is some other information that might be helpful, and I tried to proceed with the following :
Since $gcd(31,2)=1$ then is also known that $p(x)$ has only simple roots in its extension field. I know that the extension field of $x^{2^5}-x$ is $GF(2^5)$, and the roots of the irreducible polynomials (the minimal polynomials) are the groups of the conjugates (each polynomial has a group of 5/1 conjugates that are the roots of it). Now if I could say that roots of $p(x)$ in its extension field are also roots of $x^{31}-1$ since we assumed $p(x)|x^{31}-1$ (but I don't as I don't know the connection between $p(x)$ extension field and the extension field of $x^{31}-1$). So if I could say that, then because of the fact that the conjugates of the roots are also roots it must be that $p(x)$ have more roots than 8 (it implies from a theorem that says that if polynomial over a small field has root in the extension field, all of the conjugates are also roots).