Articles of primitive roots

If $r$ is a primitive root of odd prime $p$, prove that $\text{ind}_r (-1) = \frac{p-1}{2}$

If $r$ is a primitive root of odd prime $p$, prove that $\text{ind}_r (-1) = \frac{p-1}{2}$ I know $r^{p-1}\equiv 1 \pmod {p} \implies r^{(p-1)/2}\equiv -1 \pmod{p}$ But some how I feel the question wants me prove the result using indices properties (w/o factoring.. ) So I am posting it here to see if there are […]

$p^2$ misses 2 primitive roots

When I Checked primitive roots of some primes P, I found this following phenomenon: $14$ is a primitive root of prime $29$, but it’s not primitive root of $29^2$ $18$ is a primitive root of prime $37$, but it’s not primitive root of $37^2$ $19$ is a primitive root of prime $43$, but it’s not […]

How to prove 2 is a primitive root mod 37, without calculating all powers of 2 mod 37?

How can i prove 2 is a primitive root mod 37, without calculating all powers of 2 mod 37?

2 is a primitive root mod $3^h$ for any positive integer $h$

It’s easy to verify that 2 is a primitive root mod $3^2$. But then why does it follow that 2 is a primitive root mod $3^h$ for any positive integer $h$? This was used in the solution of 2009 Putnam B6 http://math.hawaii.edu/home/pdf/putnam/Putnam_2009.pdf I saw this Primitive roots of odd primes but unfortunately I don’t have […]

If $p$ and $q = 2p + 1$ are both odd primes, show that $-4$ and $2(-1)^{(1/2)(p-1)}$ are both primitive roots modulo $q$.

If $p$ and $q = 2p + 1$ are both odd primes, show that $-4$ and $2(-1)^{(1/2)(p-1)}$ are both primitive roots modulo $q$. I cannot get heads nor tails of how to even start this let alone finish it

Primitive roots modulo primes congruent to n!

for $N \ge 4$. Show for prime numbers, $p \equiv 1$ mod $(N!)$ that none of the numbers $1,2,…,N$ are primitive roots modulo $p$ I can’t figure out where to start with this question, all I can think to use is the Legendre symbol and Euler’s Criterion but I haven’t been able to do it. […]

integer $m$ has primitive root if and only if the only solutions of the congruence $x^{2} \equiv 1 \pmod m$ are $x \equiv \pm 1\pmod m$.

Show that the integer $m$ has primitive root if and only if the only solutions of the congruence $x^{2} \equiv 1 \pmod m$ are $x \equiv \pm 1\pmod m$. I don’t quite understand what this question is asking for, I thought primitive roots had order $\phi(m)$.

Prove if $n$ has a primitive root, then it has exactly $\phi(\phi(n))$ of them

Prove if $n$ has a primitive root, then it has exactly $\phi(\phi(n))$ of them. Let $a$ be the primitive root then I know other primitive roots will be among $\{a,a^2,a^3 \cdots\cdots a^{\phi(n)} \}$ because any other number will be congruent modulo $n$ to one of these. Then I figured as the answer is $\phi(\phi(n))$ so […]

Count of lower and upper primitive roots of prime $p \equiv 3 \bmod 4$

I was exploring the layout of primitive roots of primes over a reasonable range and this question concerns the number of primitive roots either side of $p/2$. Many primes have an exact match between what I call lower and upper primitive roots, those below and above $p/2$. For the most part, these are the primes […]

Proof of existence of primitive roots

In my book (Elementary Number Theory, Stillwell), exercise 3.9.1 asks to give an alternative proof of the existence of a primitive root for any prime. Let $p$ be prime, and consider the group $\mathbb{Z}/p\mathbb{Z}$. Suppose that the non-zero elements $\text{mod}\ p$ have maximum order $n < p – 1$. Show that this implies $x^n \equiv […]