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0 votes
1 answer
45 views

Proof of identity on sum of powers of primitive root.

Let $q = p^e$ for some prime $p$. Consider the trace function $\mathrm{Tr}_{\mathbb{F}_q/\mathbb{F}_p}:\mathbb{F}_q\to \mathbb{F}_p$ defined by $\mathrm{Tr}_{\mathbb{F}_q/\mathbb{F}_p}(x) = \sum_{i=0}^...
PTrivedi's user avatar
  • 1,011
2 votes
0 answers
85 views

Cyclotomic Polynomials and The Existence of Infinite Prime Power

Prove that there exist infinitely many positive integers n such that all prime divisors of $n^2 + n + 1$ are not greater than $\sqrt{n}$ This is a problem related to cyclotomic polynomial. It is ...
FaranAiki's user avatar
  • 297
2 votes
1 answer
53 views

How to prove that $\sum_\limits{k=1\\(k,p-1)=1}^{p-1}g^k \equiv \mu(p-1)$ (mod p) for prime p and primitive root g

p is a prime and g is a primitive root modules p, and I want ot prove that: $\sum_\limits{k=1\\(k,p-1)=1}^{p-1}g^k \equiv \mu(p-1)$ (mod p) $\mu(x)$ is the Möbius function I know how to deal with $\...
Gang men's user avatar
  • 425
1 vote
0 answers
314 views

Prove that the roots of cyclotomic polynomial $\Phi_{p-1}(x) \equiv 0 (mod~p)$ are exactly the primitive roots mod p

$p$ is a prime, and $\Phi_{p-1}(x)$ denote the cyclotomic polynomial of order $p-1$. And I want to show the following: $g$ is a solution of the congruence $\Phi_{p-1}(x) \equiv 0 (mod~p)$ if and only ...
Gang men's user avatar
  • 425
2 votes
1 answer
108 views

Calculation of generalized Artin's constants

Let $T(p)$ be the period of the decimal expansion of $1/p$, for prime $p$ (e.g. $1/7=0.\overline{142857}\rightarrow T(7)=6$). It is known that $$T(p)=\frac{p-1}{t}$$ for some integer $t$. Then, Artin'...
ordptt's user avatar
  • 1,069
4 votes
0 answers
161 views

Given an odd prime $p$, is there another odd prime $q$ such that $p$ is a primitive root modulo all powers of $q$?

As the title says, I want to know if for every odd prime $p$, there is another odd prime $q$ such that $p$ is a primitive root modulo $q^m$ for all $m\ge1$. For small $p$ such as $p=3,5,7$, I could ...
Pranay's user avatar
  • 271
1 vote
1 answer
43 views

Do we have an upper bound for Artin’s conjecture on primitive roots?

Let $a$ be an appropriate integer and $\pi_a (x)$ denote the number of prime $p$ such that $a$ is a primitive root modulo $p$. Do we have an upper bound of $\pi_a(x)$ such as $\pi_a(x) \ll x/\log x$? ...
yuu's user avatar
  • 101
0 votes
1 answer
68 views

Are primes of the form $6k+1$ a cube modulo $n$, if $3\nmid n$ and none of the prime factors of $n$ is of the form $6k+ 1$?

I wonder if we can assume the following statement to be true in general: Let $p$ be a prime of the form $6k+1$ and $n<p$ a natural number less than $p$. If $3$ does not divide $n$ and none of the ...
Eldar Sultanow's user avatar
4 votes
0 answers
61 views

Probability of a prime $p=3\pmod 4$ occurring in A213052

As you may notice, A213052 contains primes mostly congruent to $1\pmod 4$ (in fact, all of the known ones are except $3$). Consider the sequence of smallest primes $p_n$ such that $2,3,5,7,11,13,...$ (...
T. Rex's user avatar
  • 405
7 votes
0 answers
160 views

Intuition behind this strange heuristic for primitive roots modulo $p$?

Let $p$ be an odd prime. Define $S(p)$ as the sum of all primitive roots modulo $p$ taken from $\left[-\frac{p-1}2,\frac{p-1}2\right]$. Now here's the strange thing. If the primitive roots were '...
Mastrem's user avatar
  • 8,421
1 vote
0 answers
176 views

For what primes is $6$ a primitive root?

While thinking about an unrelated problem, I had to decide whether $6$ was a primitive root with respect to multiple prime moduli. I could discover no obvious pattern as to primes for which $6$ is a ...
Keith Backman's user avatar
-2 votes
1 answer
2k views

How to find primitive root modulo of 23? [duplicate]

These types of questions are repeated here zillionth time, but I am yet to find an useful process(hit and trial or any other process) to find primitive root modulo. Can you help me. I need this for ...
Team B.I's user avatar
  • 199
0 votes
0 answers
110 views

Why it is sufficient to look at prime divisor of $p-1$ when finding generators of $\mathbb{Z}_p^*$?

Let's say that I want to find the generators of $\mathbb{Z}_p^*$, where $p$ is a prime number. I found the following necessary and sufficient condition: An element $x \in \mathbb{Z}_p^*$ is a ...
Bean Guy's user avatar
  • 321
1 vote
1 answer
379 views

To prove that it is possible to find a prime number q with a primitive root r such that ind(p) is a prime, for a particular prime p.

Question: Let $p$ be a prime number. Prove that there exist a prime number $q$ such that for every integer $n$, the number $n^{p}-p$ is not divisible by $q$.I was trying to solve the above question ...
Pravimish's user avatar
  • 641
0 votes
1 answer
88 views

Prime of the form $n^2-2m^2$

I need to prove that a prime $p$ is of the form $n^2-2m^2$ iff $p=2$ or $p=8k\pm 1$. So first I tried figuring out, for which $p$ does $n^2-2m^2=0$ has a solution if $\mathbb{F}_p$, or similarly all ...
Math101's user avatar
  • 4,653

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