I am having trouble remembering how find the generators of a field.
Let's say I have $\Bbb{F}_5$ adjoined $\sqrt{2}$, and I want to show that $2+\sqrt{2}$ is a generator in F^x. So it's order is 34 and |F|=25. For some reason I was thinking I raise $2+\sqrt{2}$ to every power that is not co-prime to $24$, and if it not equal to $1$, it is a generator. But that is going to be awful, and there must be an easier way to accomplish this.
You know that $G=F^\times$ is a cyclic group of order $24$. Thus to show that an element $a$ is a generator, it suffices to show that $a^{12}\neq1$ and $a^8\neq 1$.
Why? Because the order of $a$ in $G$ must be a divisor of $24$. If a divisor of $24$ divides neither $12$ nor $8$, then it can only be $24$ itself.
Hence the task is simply to calculate $a^{12}$ and $a^8$.
This can be done more efficiently than multiplying $11$ times by $a$:
$$a^2=a\times a$$ $$a^4=a^2\times a^2$$ $$a^8=a^4\times a^4$$ $$a^{12}=a^8\times a^4$$
Only $4$ multiplications are needed.