TITLE: Factoring $x^{255} -1 $ over $\Bbb F_2$
QUESTION [6 upvotes]: How would I factor the above polynomial in this binary field? We just completed a course in Galois Theory, and I'm stuck on how to efficiently factor this polynomial. I tried considering computing all the irreducible polynomials of a certain degree in $\Bbb F_2$, but it is only $x^{256}-x$ in $\Bbb F_{2^8}[x]$ that is the product of all irreducible monic polynomials in the field, not the above polynomial in $\Bbb F_2$, so I'm really stuck. A tip as to the right direction to proceed would be a great answer.
Edit
I've realized that it is in fact the right choice to compute the number of monic irreducible polynomials of degree $d|n$ where $n = 256$. I will proceed from here.
Edit 2

REPLY [2 votes]: For a start, because $255=3\cdot 5 \cdot 17$ you have a difference of cubes, a difference of 5th powers and a difference of 17th powers.  So you can divide out $\sum_{i=0}^k x^i$ for $k=2,4,16$ and $x-1$.  In fact, feeding it to Alpha and ignoring the $\mathbb F_2$ gets factors with all coefficients $\pm 1$.  It splits the rest into three factors.