1. ## Primitive Roots

Show that if g and h are primitive roots of an odd prime p, then their product gh is not a primitive root of p.

2. Originally Posted by tarheelborn
Show that if g and h are primitive roots of an odd prime p, then their product gh is not a primitive root of p.

Hint: $\displaystyle x$ is a primitive root of p iff $\displaystyle x^{\frac{p-1}{2}}=-1$

Tonio

3. Originally Posted by tonio
Hint: $\displaystyle x$ is a primitive root of p iff $\displaystyle x^{\frac{p-1}{2}}=-1$

Tonio
That is not true. For instance $\displaystyle -1$ is certainly not a primitive root for any prime $\displaystyle p>3$, but $\displaystyle (-1)^{(p-1)/2}=-1$ for primes of the form $\displaystyle p=4n+3$...

What you mean is that if $\displaystyle x$ is a primtive root then $\displaystyle x^{(p-1)/2}=-1$, i.e. if $\displaystyle x$ is a primitive root then it's not a quadratic residue. Since the product of two nonresidues is a residue...

4. I am sorry, but I am not following you. Can you please elaborate?

5. If $\displaystyle x,y$ are primitive roots, then $\displaystyle x^\frac{p-1}{2}=y^\frac{p-1}{2}=-1$, and $\displaystyle (xy)^\frac{p-1}{2}=(-1)^2=1$, therefore $\displaystyle xy$ is not a primitive root.

6. Originally Posted by Bruno J.
That is not true. For instance $\displaystyle -1$ is certainly not a primitive root for any prime $\displaystyle p>3$, but $\displaystyle (-1)^{(p-1)/2}=-1$ for primes of the form $\displaystyle p=4n+3$...

What you mean is that if $\displaystyle x$ is a primtive root then $\displaystyle x^{(p-1)/2}=-1$, i.e. if $\displaystyle x$ is a primitive root then it's not a quadratic residue. Since the product of two nonresidues is a residue...

Of course. Gross typo there: if $\displaystyle w$ is a primitive root then $\displaystyle w^{\frac{p-1}{2}}=-1$. Thanx

Tonio