Why is ord$_{2^k}(5)= \frac{\varphi(2^k)}{2}$?

Question

I was doing number theory in Algebra and I came across the following example:

$$\operatorname{ord}_{2^k}(5)= \frac{\varphi(2^k)}{2} = 2^{k-2}$$

I don't understand why that is. There are no limitations set on $k$, but I suppose it doesn't bring much to consider $k<2$.

Thanks in advance!

Edit: The group in question is $(\mathbb{Z}/2^k\mathbb{Z})^*$.

Edit: As J.W.Tanner mentioned in the comment below, this works only for $k>1$. I would still like to know why.

Does this answer your question? The structure of the group $(\mathbb{Z}/2^n\mathbb{Z})^*$ — Anne Bauval, Jan 22 '23 at 20:00
@AnneBauval kinda, but what benefits does the induction in the answer have in comparison to the induction that $5^{2^{k-1}} \equiv 1$ mod $2^k$? — mikasa, Jan 23 '23 at 10:20
The more precise induction proves that the order is exactly what is wanted, and not a divisor. — Anne Bauval, Jan 23 '23 at 10:26

Robert Israel · Accepted Answer · 2023-01-22T19:15:17.420

1

I claim for $k \ge 2$, $5^{2^{k-2}} \equiv 2^k + 1 \mod 2^{k+1}$. For $k=2$ it is true: $5^{1} = 1 \cdot 2^2 + 1$. If true for $k$, then for some $a = 0$ or $1$ $$ 5^{2^{k-1}} \equiv (a 2^{k+1} + 2^k + 1)^2 \equiv 2^{k+1} + 1 \mod 2^{k+2}$$

edited Jan 22 '23 at 19:15

answered Jan 22 '23 at 17:20

Robert Israel

448,999

Wouldn't it be easier to do the induction that $5^{2^{k-1}} \equiv 1$ mod $2^k$? – mikasa Jan 22 '23 at 17:35
An extra dot appears after the first congruence symbol in this answer – J. W. Tanner Jan 22 '23 at 18:03
1

@J.W.Tanner Fixed it, thanks – Robert Israel Jan 22 '23 at 19:15
I found a duplicate – Anne Bauval Jan 22 '23 at 20:00

Why is ord$_{2^k}(5)= \frac{\varphi(2^k)}{2}$?

1 Answers1