Primitive Roots

**tarheelborn** · March 24th 2010, 09:18 AM

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.

**tonio** · March 24th 2010, 10:16 AM

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

**Bruno J.** · March 24th 2010, 12:16 PM

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...

**tarheelborn** · March 24th 2010, 12:27 PM

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

**Bruno J.** · March 24th 2010, 12:38 PM

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.

**tonio** · March 24th 2010, 01:15 PM

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

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