Order of 5 in Units of Ring of Integers Modulo 2^n
This article needs to be linked to other articles. In particular: crucial concept You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code. |
Theorem
Let $n \in \Z_{\ge 0}$ be a positive integer and $n \ge 2$.
Let $\struct {\Z / 2^n \Z, +, \times}$ be the ring of integers modulo $2^n$.
Let $U = \struct {\paren {\Z / 2^n \Z}^\times, \times}$ denote the group of units of $\struct {\Z / 2^n \Z, +, \times}$.
The order of $5$ in $U$ is $2^{n - 2}$.
Proof 1
This article needs to be tidied. Please fix formatting and $\LaTeX$ errors and inconsistencies. It may also need to be brought up to our standard house style. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Tidy}} from the code. |
This article needs to be linked to other articles. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code. |
Proof by induction:
Induction Hypothesis
For all $k \ge 3$, we can write
\(\text {(1)}: \quad\) | \(\ds 5^{2^{k - 2} }\) | \(=\) | \(\ds 1 + a_k \cdot 2^k\) |
where $a_k$ is odd.
Basis for the Induction
For $n = 3$, it means $5 ^2 = 1 + 3 \cdot 8$
Induction Step
This is our induction step:
Suppose for some $k \ge 3$, (1) is satisfied.
Consider $n = k + 1$.
\(\ds 5^{2^{k - 1} }\) | \(=\) | \(\ds \paren{5^{2^{k - 2} } }^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {1 + a_k \cdot 2^k}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + a_k 2^{k + 1} + a_k^2 \, 2^{2 k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \underbrace {a_k \paren {1 + a_k 2^{k - 1} } }_{\text{odd} } 2^{k + 1}\) |
This proves the inductive step.
Hence the result by induction.
$\Box$
Now we prove the order of $5$ in $U$ is $2^{n - 2}$.
By (1)
- $5^{2^{n - 2}} \equiv 1 \pmod {2^n}$
so the order of $5$ in $U$ divides $2^{n - 2}$.
Aiming for a contradiction, suppose it is less than $2^{n - 2}$, then it divides $2^{n - 3}$, so:
- $5^{2^{n - 3}} = 1 + m \cdot 2^n$
for some integer $m$. Then:
- $5^{2^{n - 2}} = \paren{1 + m \cdot 2^n}^2 = 1 + \paren{2 m + m^2 2^n} 2^n$
where the coefficient $2 m + m^2 2^n$ is even, contradicting lemma.
$\blacksquare$
Proof 2
Lemma
For all $k \ge 3$,
- $5^{2^{k - 3}} \equiv 1 + 2^{k - 1} \pmod{ 2^k}$
Proof of lemma
Write $\map {\nu_2} n$ for the highest exponent of a power of $2$ dividing $n$.
For all $i \ge 2$,
- $i \le 2^{i - 1}$
so
- $\map {\nu_2} i \le i - 1$
so
\(\ds \map {\nu_2} i\) | \(\le\) | \(\ds i - 1\) | ||||||||||||
\(\text {(1)}: \quad\) | \(\ds \) | \(\le\) | \(\ds 2i - 3\) | By $i \ge 2$ |
From definition of binomial coefficient:
- $\ds \binom {2^{k - 3} } i = \frac {2^{k - 3} \binom {2^{k - 3} - 1} {i - 1} } i$
we have
\(\ds \map {\nu_2} {\binom {2^{k - 3} } i}\) | \(=\) | \(\ds \map {\nu_2} {2^{k - 3} \binom {2^{k - 3} - 1} {i - 1} } - \map {\nu_2} i\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds k - 3 - \map {\nu_2} i\) | ||||||||||||
\(\text {(2)}: \quad\) | \(\ds \) | \(\ge\) | \(\ds k - 2i\) | By (1) |
so
\(\ds 5^{2^{k - 3} }\) | \(=\) | \(\ds \paren {2^2 + 1}^{2^{k - 3} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + 2^{k - 1} + \sum_{i \mathop \ge 2}\binom {2^{k - 3} } i 2^{2 i}\) | By binomial theorem | |||||||||||
\(\ds \) | \(\equiv\) | \(\ds 1 + 2^{k - 1} \pmod {2^k}\) | By (2) |
$\Box$
By the lemma,
- $5^{2^{n - 3}} \not \equiv 1 \pmod{ 2^n}$
so the order of $5$ in $U$ does not divide $2^{n - 3}$.
Squaring both sides of the lemma:
- $5^{2^{n - 2}} \equiv \paren{1 + 2^{n - 1}}^2 \equiv 1 \pmod{ 2^n}$
so the order of $5$ in $U$ divides $2^{n - 2}$.
Hence the order of $5$ in $U$ is $2^{n - 2}$.
$\blacksquare$
Sources
- Bernard (https://math.stackexchange.com/users/202857/bernard), Prove that $\text{ord}_{2^k}5=2^{k-2}$ where $k$ is any integer $\geq3$, URL (version: 2016-10-05): https://math.stackexchange.com/q/1486704