# Primitive Root Question

• Oct 27th 2010, 08:55 PM
Janu42
Primitive Root Question
Let r be a primitive root of the prime p with p congruent to 1 (mod 4). Show that -r is also a primitive root.
• Oct 28th 2010, 01:54 AM
PaulRS
Note that $\displaystyle x=r^{\tfrac{p-1}{2}} \equiv -1 (\bmod. p)$ since $\displaystyle x^2 \equiv 1 (\bmod. p)$ and so -since p is prime- $\displaystyle x\equiv 1 (\bmod. p)$ or $\displaystyle x \equiv -1 (\bmod. p)$, but the former is not possible since r is a primitive root.

Thus $\displaystyle -r \equiv r^{\frac{p+1}{2}}(\bmod. p)$ so your problem is equivalent to showing that $\displaystyle \displaystyle\frac{p+1}{2}$ is coprime to $\displaystyle p-1$.

But $\displaystyle 2\cdot \left(\displaystyle\frac{p+1}{2}\right) - (p-1) = 2$ and so $\displaystyle \text{gcd}\left(\tfrac{p+1}{2}, p-1\right)$ is either 1 or 2.

Now note that 4 doesn't divide p+1, since p-1 is divisible by 4, and so (p+1)/2 must be odd.
• Mar 22nd 2013, 03:33 AM
userit8
Re: Primitive Root Question
Quote:

Originally Posted by PaulRS
Note that $\displaystyle x=r^{\tfrac{p-1}{2}} \equiv -1 (\bmod. p)$ since $\displaystyle x^2 \equiv 1 (\bmod. p)$ and so -since p is prime- $\displaystyle x\equiv 1 (\bmod. p)$ or $\displaystyle x \equiv -1 (\bmod. p)$, but the former is not possible since r is a primitive root.

Thus $\displaystyle -r \equiv r^{\frac{p+1}{2}}(\bmod. p)$

why $\displaystyle -r \equiv r^{\frac{p+1}{2}}(\bmod. p)$ is true?