Degree of Splitting field of $x^{616} - 1$ over $\mathbb{F}_{2}[x]$

Question

How do I find the degree of the splitting field of $x^{616} - 1$ in $\mathbb{F}_{2}[x]$? I know that $x^{616} - 1 = (x^{77} - 1)^{8}$. Also, $x^{77} - 1 = (x-1) \cdot (\sum_{i=0}^{76} x^{i})$. I guess it then suffices to check the splitting field of $\sum_{i=0}^{76} x^{i}$ over $\mathbb{F}_{2}[x]$. My guess is that it is $\phi(77) = 60$, but I don't know how to show it. By Euler's theorem, $2^{60} \equiv 1 \pmod{77}$. Also, $x^{77} - 1 = \phi_{1}(x) \cdot \phi_{7}(x) \cdot \phi_{11}(x) \cdot \phi_{77}(x)$, where $\phi_{n}(x)$ is the $n^{th}$ cyclotomic polynomial. Any help would be appreciated.

Answer 1

Hint : Notice that $$x^{n}-1 = \prod\limits_{d \mid n} \phi_{d}(x)$$

where $\phi_{m}(x)$ is the $m$-th cyclotomic polynomial, i.e. $\phi_{m}(x) = \prod\limits_{i=1}^{\phi(m)} (x-\alpha_{i})$, with $\alpha_{i}$ beeing the $m$-th primitive roots of unity.

It's a known result the following :

Theorem 1 : Let $p \nmid n$. The irreducible factors of $\bar{\phi}(x) \in \mathbb{F}_{p}[x]$ are distinct and their degrees are equal to the order of $p$ in $(\mathbb{Z}_{n})^{*}$

Let's take $p(x) = x^{77}-1 = \phi_{1}(x) \phi_{7}(x) \phi_{11}(x) \phi_{77}(x)$.

Also notice that the splitting field is equal to $\mathbb{F}_{p^k}$, where $k$ is the lcm of the degrees of the irreducibles factors of the given polynomial.

Can you take it from here ?