Questions tagged [multiplicative-order]
Let $G$ be a finite group, typically $\mathbb{Z}/n \mathbb{Z}$, and $g\in G$. The multiplicative order of $g$ is the least $n\in\mathbb{N}^+$ such that $g^n = e$, the identity of $G$.
72 questions
5 votes
1 answer
248 views
+50
Primes for which sum of sine ratios is nearly always an integer
Let $p$ be a prime, $1\le b \le p-1$ be an integer, and $\text{ord}(b)$ be the order of $b$ mod $p$. I am interested in the sum $$S_p(b) = \sum_{k=1}^{\text{ord}(b)}\frac{\sin\left(b^{k+1}\cdot\frac{p-...
0 votes
0 answers
47 views
Common element with order constraints
Let $$2^{\ell-1}<p_1<p_2<\dots<p_t<2^\ell$$ $$2^{\ell'-1}<q_1<q_2<\dots<q_m<2^{\ell'}$$ be primes on the condition $$\phi(p_i)=2q_1^{a_1}\dots q_m^{a_m}$$ ($\phi$ is ...
0 votes
0 answers
59 views
For coprime $m,n$, $\operatorname{ord}_{mn} (a) = \operatorname{lcm} (\operatorname{ord}_n (a), \operatorname{ord}_m (a))$ [duplicate]
To make sure I'm not going crazy, since this is surprisingly difficult to find online: $\DeclareMathOperator{\ord}{ord}$ Let $\ord_n (a)$ be the minimum $r > 0$ with $a^r \equiv 1 \mod n$ (when it ...
1 vote
0 answers
131 views
Is there always a prime that divides $10^k +1$, but does not divide any $10^n +1 ~, ~ n < k?$
I was looking at numbers of the form $$ N(k) = 10^k + 1 $$ and their factors. I noticed for some small $k$, it seems that $10^k + 1$ always has at least one prime factor that is not shared by any $N(n)...
0 votes
1 answer
63 views
Proof verification : $1+p\in (\mathbb Z/p^n\mathbb Z)^\times$ and its order is $p^{n-1}$
I want to show that $1+p$ is an element of $(\mathbb Z/p^n\mathbb Z)^\times$ and its order is $p^{n-1}$ for odd prime $p$. (Dummit/Foote 2.3.21) The textbook said one can prove it by using the ...
2 votes
0 answers
90 views
number of primitive prime divisor
Let $q = p^n$ for some prime number $p > 2$. According to Zsigmondy's theorem, for the number $q - 1$ there is at least one prime primitive divisor, that is, a number $r$ such that $\operatorname{...
6 votes
0 answers
81 views
Equivalence of multiplicative order modulo prime [duplicate]
Given $2$ positive integers $a, b$. If for each prime $p$ such that $p\nmid a$ and $p\nmid b$, the multiplicative order of $a$ modulo $p$ always equal the multiplicative order of $b$ modulo $p$, does ...
0 votes
2 answers
109 views
What is the order of $p_{1}^{x} \bmod{n}$ where $p_1$ is a prime factor of $n$ [closed]
I am looking for a formula, algorithm, or even literature on the topic. Take $21$ for example $21 = 7 \cdot 3$ What is the order of $3^{x} \bmod 21$? $3^0 = 1$ $3^1 = 3$ $3^2 = 9$ $3^3 = 6$ $3^4 = 18$...
1 vote
1 answer
85 views
Properties of matrices with a multiplicative order
I have been trying to find any article or sources talking about the structure and properties of matrices with a multiplicativw order, i.e. a matrix $A$ has a multiplicative order of $n$ if and only if ...
1 vote
0 answers
60 views
Infinite sequence of integers $\{m_n\}$ for which the orders of both 2 and 3 are small modulo $m_n$.
It is easy to create a sequence $\{m_n\}$ for which the order of $2\pmod{m_n}$ is as small as possible, i.e. it is about $\log_2(m_n)$. For example $m_n=2^n-1$ is an appropriate sequence. But if I ...
0 votes
1 answer
210 views
n = pq where p and q are odd prime numbers, gcd(c,n) = 1, c < n, show at most $\frac{φ(n)}{4}$ of c satisfy $ord_n(c)$ is odd
Suppose n = pq where p and q are distinct odd prime numbers. Show that, out of the φ(n) different integers c satisfying 1 < c < n and gcd(c,n) = 1, at most $\frac{φ(n)}{4}$ of them have the ...
3 votes
0 answers
199 views
Why is the multiplicative order of perfect powers modulo $p$ smaller on average?
I've been trying to analyse by myself for recreational purposes what would be a "better" base to use instead of the common decimal one. Part of what should make a base better is to have ...
0 votes
4 answers
116 views
$m+1$ generates the kernel of $Z_n^\times\to Z_m^\times$ where $m\mid n$ with the same prime factors
Suppose $m\mid n$. Using the First Isomorphism Theorem with respect to the homomorphism $$\begin{array}{rccc}f:&\mathbb{Z}_n^\times&\to&\mathbb{Z}_m^\times \\&x&\mapsto &x\bmod ...
3 votes
0 answers
87 views
About order of a number modulo 2. [duplicate]
For $a, m, n \in \mathbb{N}$, prove that $$\gcd(a^{2^m} + 1, a^{2^n} + 1) = \begin{cases} 1, & \text{if $a$ is even}\\ 2, & \text{if $a$ is odd} \end{cases}$$ given $m \ne n$. My attempt: ...
2 votes
1 answer
1k views
Multiplicative order of $2$ modulo $p$.
When calculating the multiplicative order of $2$ modulo a prime $p$ you often get $p-1$ or $\frac{p-1}{2}$ as a result, but there are cases where this does not hold, is there a general form for those ...