find the condition for k such that is also a primitive root mod 13.
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.
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 is a primitive root of , all the primitive roots are given by , where ; so the primitive roots of are and .
Then the product of all the primitive roots of is congruent to modulo . By Fermat's Theorem, .
The general result is the following: Let be a positive integer which has a primitive root. Then the product of the primitive roots of is congruent to modulo ; the exceptions are and .
By the way, the integers that have primitive roots are and , where is an odd prime.
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?