Order of 5 in Units of Ring of Integers Modulo 2^n

From ProofWiki
Jump to navigation Jump to search


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.



This page has been identified as a candidate for refactoring of basic complexity.
Until this has been finished, please leave {{Refactor}} in the code.

New contributors: Refactoring is a task which is expected to be undertaken by experienced editors only.

Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$.
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 {{Refactor}} 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

\(\ds 5^{2^{k - 3} }\) \(=\) \(\ds \paren {2^2 + 1}^{2^{k - 3} }\)
\(\ds \) \(=\) \(\ds 1 + 2^{k - 1} + \sum_{i \mathop \ge 2}\binom {2^{k - 3} } i 2^{2 i}\) Binomial Theorem for Integral Index
\(\ds \) \(\equiv\) \(\ds 1 + 2^{k - 1} \pmod {2^k}\) 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 $2^{n - 2}$.

$\blacksquare$

Generalization


This page or section has statements made on it that ought to be extracted and proved in a Theorem page.
In particular: Put this in its own page
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by creating any appropriate Theorem pages that may be needed.
To discuss this page in more detail, feel free to use the talk page.


Let $U$ be the same as above.

Let $u$ be any odd number of the form $8k + 3$ or $8k + 5$.

The order of $u$ in $U$ is $2^{n - 2}$.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Order_of_5_in_Units_of_Ring_of_Integers_Modulo_2%5En&oldid=738930"