- #1

- 14

- 0

I've been told it's quite easy, but I can't see why it's true for the life of me

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- Thread starter RichardCypher
- Start date

- #1

- 14

- 0

I've been told it's quite easy, but I can't see why it's true for the life of me

- #2

- 1,056

- 0

If r is a primitive root, then r^[(p-1)/2] is not equal to 1. The question is about -r.

- #3

- 14

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if r is a primitve root then r^[(p-1)/2] is -1 (mod p). However, (-r)^[(p-1)/2] is the same since (p-1)/2 is even, am I right?

Where do we go from here? Maybe I could say that the order of (-r) couldn't be lower than (p-1), because if it was so, ord(-r)|(p-1)/2 and then (-r)^[(p-1)/2]=1(mod p) contradicting the fact that p [tex]\neq[/tex] 2?

Am I on the right track?

Thanks again!

- #4

- 1,056

- 0

Sounds O.K. to me. Take, (-r)^2 = r^2, well then what power is necessry to raise r^2 to 1?

- #5

- 14

- 0

OK, I got it!

Bunch of thanks for your help!

Bunch of thanks for your help!

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