TITLE: Show that there are infinitely many reducible polynomials of the form $x^n+x+1$ in $\mathbf{F}_2[x]$
QUESTION [6 upvotes]: Here is a question from an old exam:

Show that there are infinite $n\in \mathbf{N}, A= x^{n}+x+1 $ which are reducible over  $\mathbf{F}_{2}[x]$. 

Using André Nicolas' and Qiaochu Yuan's hint: $x^{2}+x+1$ as dividing polynomial. $x^{2}+x+1$ is irreducible over $\mathbf{F_{2}}$. If an irreducible polynomial divides another polynomial which is not itself, that means that polynomial must be reducible. We want to show that $x^{2}+x+1$ divides all polynomials of the form $x^{3n+5}+x+1$. I can't figure the induction steps, but in $\mathbf{F_{2}}$ the polynomial belongs to the residue class $\tilde{1}$, therefore there must be an infnite amount of them.
Concerning Gerry Myerson's hint, how can I use cubic roots in $\mathbf{F_{2}}$, wouldn't I need $\mathbf{R}[i]$ for that? 
Help is greatly appreciated.

REPLY [9 votes]: As a variation on the (essentially equivalent) ideas in the answers and comments: we could ask that there be $\alpha\in \mathbb F_4$ solving $x^n+x+1=0$, noting that certainly there is no solution in $\mathbb F_2$. For $\alpha\in \mathbb F_4$, $\alpha^4=\alpha$, and since $\alpha\not=0,1$, also $\alpha^3=1$ and $\alpha^2+\alpha+1=0$. Thus,
$$
\alpha^{3n+2} + \alpha + 1 = \alpha^2+\alpha+1 = 0
$$
Thus, $x^2+x+1$ divides $x^{3n+2}+x+1$.