# Primitive Roots

• March 24th 2010, 09:18 AM
tarheelborn
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.
• March 24th 2010, 10:16 AM
tonio
Quote:

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

Tonio
• March 24th 2010, 12:16 PM
Bruno J.
Quote:

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

Tonio

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

What you mean is that if $x$ is a primtive root then $x^{(p-1)/2}=-1$, i.e. if $x$ is a primitive root then it's not a quadratic residue. Since the product of two nonresidues is a residue...
• March 24th 2010, 12:27 PM
tarheelborn
I am sorry, but I am not following you. Can you please elaborate?
• March 24th 2010, 12:38 PM
Bruno J.
If $x,y$ are primitive roots, then $x^\frac{p-1}{2}=y^\frac{p-1}{2}=-1$, and $(xy)^\frac{p-1}{2}=(-1)^2=1$, therefore $xy$ is not a primitive root.
• March 24th 2010, 01:15 PM
tonio
Quote:

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

What you mean is that if $x$ is a primtive root then $x^{(p-1)/2}=-1$, i.e. if $x$ 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 $w$ is a primitive root then $w^{\frac{p-1}{2}}=-1$. Thanx

Tonio