Multiplicative Order when for integers $m > 2, n > 0$, when $2^m | (3^n - 1)$

Question

It seems to me that for integers $m>2, n>0$, when $2^m | (3^n - 1)$, that the multiplicative order is $2^{m-2}$ so that for $0 < i < 2^{m-2}$, $3^i \not\equiv 1 \pmod {2^m}$ and $3^{2^{m-2}}\equiv 1 \pmod {2^m}$

I am having trouble completing the inductive argument.

Here's what I came up with for the base case. For $m=3$, $2^3 | (3^n-1)$ if and only if $2 | n$ since $3^{2t+1} \equiv 3\times9^t \equiv 3 \pmod 8$ and $3^{2t} \equiv 9^t \equiv 1^t \pmod 8$

Assume that up to some $m \ge 3$, $2^m | (3^n - 1)$ if and only if $2^{m-2} | n$

Assume that $2^{m+1} |(3^n-1)$. It follows that there exists an integer $a$ such that $n = 2^{m-2}a$.

Assume that $a$ is odd. It seems to me that it should be possible to complete this argument through a contradiction. I am not clear how to finish the argument.

FYI, a fairly closely related question is Prove that if $n$ is a positive integer, then $3^{2^{n}}-1$ is divisible by $2^{n+2}$, although that one just asks to prove divisibility rather than it being specifically the multiplicative order. Also, there's Prove that $2^{n+2}$ is a divisor of the number $k^{2^{n}}-1$, showing the divisibility applies to all odd integers, not just $3$. — John Omielan, May 27 '21 at 07:43

John Omielan · Accepted Answer · 2022-05-30T09:18:52.610

0

Hint: Note with $n = 2^{m-2}a$ that

$$3^n - 1 = (3^{2^{m-2}} - 1)(\color{blue}{3^{2^{m-2}(a-1)} + 3^{2^{m-2}(a-2)} + \ldots + 3^{2^{m-2}} + 1}) \tag{1}\label{eq1A}$$

There are $a$ terms in $\color{blue}{\text{blue}}$. If $a$ is odd, then since each term is odd, the sum will also be odd.

edited May 30 '22 at 09:18

answered May 27 '21 at 07:18

John Omielan

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Multiplicative Order when for integers $m > 2, n > 0$, when $2^m | (3^n - 1)$

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