What is the number of irreducible factors of $x^{255}-1$ over $\mathbb{Q}$ and $\mathbb{F}_2$? Over $\mathbb{Q}$ $x^{255}-1$ factorises as $(x-1) \Phi_{5} \Phi_{51}\Phi_{255}$ with all of them irreducible but I am not sure if this is correct.
As for $\mathbb{F}_2$ I have no clue about how to proceed.
 A: Your factorization over $\Bbb{Q}$ is not quite correct; note that $255=3\times5\times17$.
For a factorization over $\Bbb{F}_2$; use the fact that $x^{p^n}=x$ for all  $x\in\Bbb{F}_{p^n}$. This means $x^{255}=1$ for all nonzero $x\in\Bbb{F}_{2^8}$, and so $x^{255}-1$ is divisible by the minimal polynomial of every element of $\Bbb{F}_{2^8}^{\times}$. So every irreducible polynomials whose degree divides $8$ is a factor of $x^{255}-1$.
A: Welcome to MSE! Hint: The polynomial $x^{p^n}-x$ in ${\Bbb Z}_p[x]$, $p$ prime, is the product of all monic irreducible polynomials in ${\Bbb Z}_p[x]$ whose degree is a divisor of $n$, i.e., $$\sum_{d|n} d N_d = p^n,$$ where $N_d$ is the number of monic irreducible polynomials of degree $d$ in ${\Bbb Z}_p$.
To be more than a comment, here is the factorization for $x^{16}-x$ in ${\Bbb Z}_2[x]$, which is part of the problem:
Degree 4: $x^4+x+1$, $x^4+x^3+1$ (conjugate to the first polynomial, the zeros of these polynomials are primitive), and $x^4+x^3+x^2+x+1$ (the zeros are 5-th roots of unity).
Degree 2: $x^2+x+1$.
Degree 1: $x$ and $x+1$.
Now to the solution:
$$256 = 30\cdot 8 + 3\cdot 4 + 1\cdot 2 + 2\cdot 1.$$
A: Over $\Bbb Q$, $x^n-1$ factors as $\prod_{d\mid n}\Phi_d(x)$
where the $\Phi_d$ are cyclotomic polynomials. These are irreducible
over $\Bbb Q$. The number of irreducible factors of $x^n-1$
over $\Bbb Q$ is the number of (positive integer) divisors of $n$.
