Showing every element in $\mathbb{F}_{25}$ is a power of a given element

Question

Let $K$ be a field of order $25$, with the basis $\{a,1\}$ as a vector space over $\mathbb{F}_5$.

Let $a^2 = 3 \text{ and } b := a + 1$.

From here it is quite easy to see that $b^{-1} = 2 + 3a$; now I need to show that every element in $K^\times$ is a power of $b$; i.e., all linear combinations $x \cdot a + y \text{, with } x,y \in \mathbb{F}_5 \wedge \left( x \neq 0 \vee y \neq 0 \right)$ are powers of $b$. Is there a smarter way to do this than brute force calculating all powers of $b$?

Can you use the fact that $K^\times$ is cyclic? – J. W. Tanner May 21 '23 at 14:32 — J. W. Tanner, May 21 '23 at 14:32
I can't use that... – strugglingStudent May 21 '23 at 14:37 — strugglingStudent, May 21 '23 at 14:37
Can you show $b^{24}=1$ but $b^{12}\ne1$ and $b^8\ne1$? – J. W. Tanner May 21 '23 at 14:50 — J. W. Tanner, May 21 '23 at 14:50
Yes, and that would help me how? – strugglingStudent May 21 '23 at 15:06 — strugglingStudent, May 21 '23 at 15:06
For the case of a different generator see here or here. – Jyrki Lahtonen May 21 '23 at 15:31 — Jyrki Lahtonen, May 21 '23 at 15:31

J. W. Tanner · Accepted Answer · 2023-05-21T15:12:46.880

The cyclic group generated by $b$ under multiplication is a subgroup of $K^\times$.

By calculating some powers of $b$, I show it has order $24$ and thus comprises the entire group.

$b:=a+1$

$b^2=(a+1)^2=a^2+2a+1=2a+4$

$b^4=(b^2)^2=(2a+4)^2=4(a+2)^2=4(a^2+4a+4)=4(2+4a)=a+3$

$b^8=(b^4)^2=(a+3)^2=a^2+6a+9=a+2$

$b^{12}=b^8b^4=(a+2)(a+3)=a^2+5a+6=4$

$b^{24}=(b^{12})^2=4^2=1$.

Since $b^{24}=1$ but $b^8\ne1$ and $b^{12}\ne1$, it follows that $b$ generates the entire group $K^\times$,

which has all the $24$ distinct linear combinations $x\cdot a+y$ with $x,y\in \mathbb F_5$, $x\ne0 $ or $y\ne0$.

Showing every element in $\mathbb{F}_{25}$ is a power of a given element

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