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[also under discussion in math links forum]

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- March 1st 2010, 04:16 PMkingwinnerPrimitive Root of odd prime p
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[also under discussion in math links forum] - March 1st 2010, 09:53 PMchiph588@
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 . - March 2nd 2010, 02:09 AMtonio
- March 2nd 2010, 09:33 AMkingwinner
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! - March 2nd 2010, 11:10 AMchiph588@
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. .