Order of elements in the multiplicative group of the finite field extension

Question

Given the field extension $\mathbb{F}_5(\sqrt2)$, it is isomorphic to $\mathbb{F}_{5^2}$ and the group $\mathbb{F}_{5^2}^{*}$ is cyclic of degree 24 so the elements in it above can be of order $\{2,3,4,6,8,12\}$. How can I calculate the orders of $1 – \sqrt{2}$, $3 – \sqrt2$ in $\mathbb{F}_{5}(\sqrt2)^{*}$ without having to go through all the possible power of said elements ?

Answer 1

You don't need to compute all powers, only a few.

Let $v=1 - \sqrt2$. Then $v^3=2$ and so $v^6=4=-1$. Therefore, $v$ has order $12$.

Let $w=3 - \sqrt2$. Then $w^3=\sqrt2$ and so $w^{12}=4=-1$. Therefore, $w$ has order $24$.