1. ## 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.

2. 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.

3. ## Re: Primitive Root Question

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?