Primitive roots and there negatives

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SUMMARY

The discussion centers on proving that if \( p \) is a prime of the form \( p=4k+1 \) and \( g \) is a primitive root of \( p \), then \( -g \) is also a primitive root. The proof demonstrates that the order of \( -g \) modulo \( p \) is \( 4k \) by assuming \( ord_{p}(-g) = d \) and showing that \( d \) cannot be even or an odd factor of \( k \), leading to contradictions. Therefore, it is established that \( -g \) is indeed a primitive root.

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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 thatord_{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)=2da contradiction.

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

Homework Statement


 
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Hi talolard! :smile:

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

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