Primitive Root

Let $n$ be an positive integer. Then integer $a$ is said to be a Primitive Root of $n$ (if it exist), if

\[\text{ord} _{n} a = \phi (n)\]

Existence

Primitive roots only exist for certain integers. In fact, it only exist for $n=2,4, p^t$ or $2p^t$, where $p$ is a ODD prime and $t$ is a positive integer.

The proof of that statement is extremely long and tedious. Euler tried to prove it but his proof was wrong. This is a link to the proof: Proof of the Existence of Primitive Roots

How to Find a Primitive Root

Once you find a primitive root for a prime number $p$ (which you probably just have to find by trial and error), $r$, determine if

\[r^{p-1} \equiv 1 \pmod {p^2}\]

If $r^{p-1} \equiv 1 \pmod {p^2}$, then $r$ is not a primitive root of $p^2$, but $r+p$ is. Now, $r^{p-1} \equiv 1 \pmod {p^2}$ is extremely rare, so for the vast majority of the time, if $r$ is a primitive root for $p$, it is also a primitive root of $p^2$. However, if $r+p$ is a primitive root of $p^2$, then obviously $r+p$ is also a primitive root of $p$. Now, if $r$ is a primitive root of both $p$ and $p^2$, then $r$ is a primitive root of $p^k$ for all positive integers $k$.


A proof of all the statements above can be found here.

Uses

Primitive roots have some important properties.

For example, if $r$ is a primitive root of a integer $n$ and $r$ and $n$ is relatively prime, then

\[S= \{ r, r^2, r^3, \dots , r^{\phi (n)} \}\]

form a reduced residue system modulo $n$.

Proof:

Obviously, every elements of $S$ is relatively prime to $n$.

Now, suppose that there exist two integers $j$ and $k$ such that $1 \le j,k \le \phi (n)$ satisfies

\[r^j \equiv r^k \pmod {n}\]

Then

\[r^{j-k} \equiv 1 \pmod {n}\]

so $j-k=0$, since $0 \le |j-k| < \phi (n)$. $\square$


This result allows us to define the index of an integer with respect to a prime $p$ and a primitive root of $p$ , $r$.

Indices can be used to solve Polynomial Congruences.

See Also

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