How many primitive elements does GF(256) have?

Question

I know the answer for this is 36 but I don't exactly know how to reach to this. Can you any one help me in understanding this.

Answer 1

The number of primitive elements in a finite field $GF(n)$ is $φ(n - 1)$, where $φ(m)$ is Euler's totient function. Now compute $\phi(255)$. Use that $255=3\cdot 5\cdot 17$.

Answer 2

An article about Primitive Polynomial in Wolfram Mathworld says,

There are $$a_q(n)=\frac{\phi(q^n-1)}{n}$$ primitive polynomials over $\mathrm{GF}(q)$, where $\phi(n)$ is the totient function.

So, the number of primitive polynomials over $\mathrm{GF}(2^8)$ is supposed to be $\phi(255)/8=16$.