Number Theory

**peteryellow** · January 10th 2009, 11:58 AM

I want to determine the smallest prime $p$ such that
$ \left(\frac {2}{p}\right)=-1 $

and the order of $[2]_p$ is less than $p-1$ .

Here $ \left(\frac {2}{p}\right)=-1 $
is Legendre symbol.

I have tried with different prime number but have not got any answer.

**o_O** · January 10th 2009, 01:12 PM

$p \equiv 3 \ \text{or} \ 5 \ (\text{mod } 8) \ \ \Rightarrow \ \ (2/p) = - 1$

With the help of Wiki's Table of Primitive Roots

, the first prime that does not have 2 as its primitive root and is congruent to 3 or 5 is $p=13$

**peteryellow** · January 11th 2009, 01:02 AM

but the order of [2]_13 is 12, and I am looking for a p where order of 2 is less than p-1?

**ThePerfectHacker** · January 11th 2009, 09:32 AM

Originally Posted by peteryellow

but the order of [2]_13 is 12, and I am looking for a p where order of 2 is less than p-1?

If you look at the table o_O provided thou shall see that $13$ has primitive root $6$ and that $[2]_{13} = [6]_{13}^5$ .
Thus, $[2]_{13}$ is not a primitive root.

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