- #1
mathmari
Gold Member
MHB
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Hey! :giggle:
How can we calculate the number of natural numbers between $2$ and $n$ that have primitive roots?
Let $m$ be a positive integer.
Then $g$ is a primitive root modulo $m$, with $(g,m)=1$, if the modulo of $g\in (Z/m)^{\star}$ is a generator of the group.
We have that $g$ is a primitive root modulo $m$ if it is a generator of a group, i.e. $m$ has a primitive root if $\mathbb{Z}_m$ is cyclic, right?
$\mathbb{Z}_m$ is cyclic if $m=1,2,4$ or $m=p^k$ or $m=2\cdot p^k$ for $p$ prime.
That means that the number of natural numbers that have a primitive root is $\#\{1,2,4,p^k, 2\cdot p^k\}$ for $p$ prime.
So we have to calculate the number of primes between $2$ and $n^{\frac{1}{k}}$ to calculate then the number of elements of the form $p^k$ and $2\cdot p^k$.
Have I understood that correctly? :unsure:
How can we calculate the number of natural numbers between $2$ and $n$ that have primitive roots?
Let $m$ be a positive integer.
Then $g$ is a primitive root modulo $m$, with $(g,m)=1$, if the modulo of $g\in (Z/m)^{\star}$ is a generator of the group.
We have that $g$ is a primitive root modulo $m$ if it is a generator of a group, i.e. $m$ has a primitive root if $\mathbb{Z}_m$ is cyclic, right?
$\mathbb{Z}_m$ is cyclic if $m=1,2,4$ or $m=p^k$ or $m=2\cdot p^k$ for $p$ prime.
That means that the number of natural numbers that have a primitive root is $\#\{1,2,4,p^k, 2\cdot p^k\}$ for $p$ prime.
So we have to calculate the number of primes between $2$ and $n^{\frac{1}{k}}$ to calculate then the number of elements of the form $p^k$ and $2\cdot p^k$.
Have I understood that correctly? :unsure: