1. ## Number Theory

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

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

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

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

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

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 $\displaystyle p=13$

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

4. 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 $\displaystyle 13$ has primitive root $\displaystyle 6$ and that $\displaystyle [2]_{13} = [6]_{13}^5$.
Thus, $\displaystyle [2]_{13}$ is not a primitive root.