# Primitive roots - annoying problem

## Main Question or Discussion Point

Let $$r$$ be a primitive root of a prime number $$p \geq 3$$. Prove that if $$p \equiv 1 (mod 4)$$, then $$-r$$ is also a primitive root of $$p$$.

I've been told it's quite easy, but I can't see why it's true for the life of me Related Linear and Abstract Algebra News on Phys.org
If r is a primitive root, then r^[(p-1)/2] is not equal to 1. The question is about -r.

thanks for the reply if r is a primitve root then r^[(p-1)/2] is -1 (mod p). However, (-r)^[(p-1)/2] is the same since (p-1)/2 is even, am I right?

Where do we go from here? Maybe I could say that the order of (-r) couldn't be lower than (p-1), because if it was so, ord(-r)|(p-1)/2 and then (-r)^[(p-1)/2]=1(mod p) contradicting the fact that p $$\neq$$ 2?
Am I on the right track?

Thanks again! Richard Cypher: Am I on the right track?

Sounds O.K. to me. Take, (-r)^2 = r^2, well then what power is necessry to raise r^2 to 1?

OK, I got it!
Bunch of thanks for your help! 