[also under discussion in math links forum]
Let be even.
Then since is a primitive root.
If is odd, then , but for this would mean which we know to be false since .
Therefore is a primitive root if .
If then , so i.e. is odd. Therefore .
Now since . Let .
We know . But since is a primitive root, .
Therefore . Hence is not a primitive root for .
Hi, thanks for your help! I have some questions about your proof.
1) When you say k|p-1 is odd or k|p-1 is even, do you include the case k=p-1? Or are you only referring to the divisors of p-1 that are strictly less than p-1?
2) Also, how do you know that 2k is strictly less than p-1? (i.e. 2k < p-1?)
Thank you very much!
1.) Yes, I meant to say k is strictly less than p-1
2.) What's the smallest divisor of p-1 not equal to 1? The answer is 2.
Therefore is the largest divisor strictly less than p-1.
But we know and is odd.
So, since and is the largest proper divisor we have
i.e. .