sophie germain prime primitive root

Here p denotes a prime.
Suppose q is a prime such that q=4n+1 where n is an interger.
proof that 2 is a primitive root of p if p is of the form 2q+1.
(sophie germain prime).

Quote:

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Originally Posted by santiagos11

Here p denotes a prime.
Suppose q is a prime such that q=4n+1 where n is an interger.
proof that 2 is a primitive root of p if p is of the form 2q+1.
(sophie germain prime).

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First, $\displaystyle \phi(p)=p-1=2q\Longrightarrow ord_q(2)\in\{1,2,q,2q\}$ . Now, 1 obviously cannot be, and $\displaystyle 2^2=1\!\!\!\pmod p\Longrightarrow p\mid 3$ , which is also impossible.

Now using Jacobi symbol, we know that $\displaystyle \left(\frac{2}{p}\right)=2^{\frac{p-1}{2}}=2^q\!\!\!\pmod p$ , but since $\displaystyle p=3\!\!\!\pmod 8\neq \pm 1\!\!\!\pmod 8$ , then $\displaystyle \left(\frac{2}{p}\right)=-1\Longrightarrow 2^q=-1\!\!\!\pmod p\Longrightarrow ord_p(2)\neq q$ , and thus...

Tonio

shouldn't this be $\displaystyle \phi(p)=p-1=2q\Longrightarrow ord_p(2)\in\{1,2,q,2q\}$. (order mod p instead of mod q)
And why is this implication true?

Quote:

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Originally Posted by santiagos11

shouldn't this be $\displaystyle \phi(p)=p-1=2q\Longrightarrow ord_p(2)\in\{1,2,q,2q\}$. (order mod p instead of mod q)

Yes. That was a typo.

And why is this implication true?

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Because $\displaystyle \{1,2,q,2q\}$ are the only numbers dividing $\displaystyle p-1=2q$ ...(Itwasntme)

Tonio

ok thanks, you know sometimes we see the hard things easily and the easy things are hard to see!