Primitive roots

**Cairo** · March 7th 2011, 10:00 AM

If g is a primitive root mod 13 then I can show that g^11 is also a primitive root.

My question is how can I deduce that the product of all the primitive roots mod 13 are congruent to 1 (mod 13) from this result?

Is there a corresponding result for the product of all the primitive roots mod 169?

Thanks in advance.

**Shanks** · March 9th 2011, 01:54 AM

find the condition for k such that $g^k$ is also a primitive root mod 13.

**Cairo** · March 9th 2011, 10:01 AM

I'm aware of the condition for k to such that g^k is a primitive root mod 13. In fact, I have shown that g^11 is a primitive root mod 13.

But my question is how can I use this information to deduce that the product of all the primitive roots mod 13 is congruent to 1 mod 13.

I was then curious if a similar result holds for the product of all primitive roots mod 169.

**melese** · March 9th 2011, 11:17 AM

Originally Posted by Cairo

I'm aware of the condition for k to such that g^k is a primitive root mod 13. In fact, I have shown that g^11 is a primitive root mod 13.

But my question is how can I use this information to deduce that the product of all the primitive roots mod 13 is congruent to 1 mod 13.

I was then curious if a similar result holds for the product of all primitive roots mod 169.

Given that $g$ is a primitive root of $13$ , all the primitive roots are given by $g^k$ , where $(k,12)=1$ ; so the primitive roots of $13$ are $g^1, g^5, g^7,$ and $g^{11}$ .

Then the product of all the primitive roots of $13$ is congruent to $g^{1+5+7+11}=g^{24}$ modulo $13$ . By Fermat's Theorem, $g^{24}=(g^{12})^2\equiv1\pmod {13}$ .

The general result is the following: Let $m$ be a positive integer which has a primitive root. Then the product of the primitive roots of $m$ is congruent to $1$ modulo $m$ ; the exceptions are $m=3, 4,$ and $6$ .

By the way, the integers $m$ that have primitive roots are $1, 2, 4, p^k,$ and $2p^k$ , where $p$ is an odd prime.

**Cairo** · March 9th 2011, 10:13 PM

Thank you both for your help with this.

I managed to prove the first result in the end. I never thought to use Fermat's Theroem until It struck me at the last minute.

I would like some clarification on your second point melese if that is possible?

If g is a primitive root mod 169, then so is g+13. So how are the rest of the primitive roots obtained and how does their product become congruent to 1 mod 169?

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