Primitive roots and there negatives

[talolard]

Homework Statement

if p is a prime of the form p=4k+1 and g is a primitive root of p, show that -g is a primitive root.


I'm not sure if this is a decent proof or not. My final argument looks suspicious. Any thoughts?
Thanks
Tal

The Attempt at a Solution

First, notive that $$\phi(p)=4k$$. we wish to show that $$ord_{p}(-g)=4k$$.

Assume that $$\left(-g\right)^{d}\equiv1(p)$$ and $$d\neq4k$$ then d divides 4k.

Assume that d=2a then $$\left(-g\right)^{2a}=1\cdot g^{2a}$$ implies that$$ord_{p}(g)=2a$$ a contradiction. Thus d must be odd.

Assume that d is an odd factor of k. then $$\left(-1g\right)^{d}=-g^{d}\equiv1(p)\iff g^{d}\equiv-1\iff g^{2d}=1 thus ord_{p}(g)=2d$$a contradiction.

Thus $$ord_{p}(-g)=4k$$ and -g is a primitive root.

Homework Statement

 

[micromass]

Hi talolard!

That looks like a decent proof to me! You may want to explain why $ord_p(g)=2d$ is a contradiction.

 

[talolard]

Great!
Thanks