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

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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


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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

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