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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 ?

endeavor
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    You don't need to compute all powers. Let $v=1 - \sqrt2$. Then $v^3=2$ and so $v^6=\cdots$. Let $w=3 - \sqrt2$. Then $w^3=\cdots$. – lhf Jun 07 '21 at 13:36
  • @lhf I understand that just by computing the 2nd and 3rd power we could obtain the rest but is there any method that does not require computing any power at all ? – endeavor Jun 07 '21 at 14:41
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    I don't know, if you are looking for something like this. You see, I'm a fan of ad hoc tricks. Those do not generalize very well :-) – Jyrki Lahtonen Jun 07 '21 at 19:13
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    It’s not a matter of “by computing the 2nd and 3rd power we could obtain the rest”, but of noting the (already known) order of the square and cube. I would also like to express the opinion that you should embrace with joy the idea of doing pencil-and-paper computations. – Lubin Jun 11 '21 at 22:06

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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 $24$.

lhf
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