a) and b) Suppose a = primitive root mod p for an odd prime p. Show that $\left(\frac{a}{p}\right)$ = -1 I understand how to evaluate Legendre symbols when given discrete numbers.
For example the Legendre symbol for the following:
$\left(\frac{3}{105953}\right)=\left(\frac{105953}{3}\right)=\left(\frac{2}{3}\right)=-1$ 
I obtained -1 using quadratic reciprocity and fact that $105953\equiv 1\bmod 4$
However, I am having trouble with the following questions:
a) For which odd primes p is $\left(\frac{-2}{p}\right) = 1$ ? 
b) Suppose a = primitive root mod p for an odd prime p. Show that $\left(\frac{a}{p}\right)$ = -1
 A: By definition, Legendre symbol $(\frac{a}{p})=1$ if $a$ is quadratic residue, -1 otherwise. Primitive root $a$ cannot be quadratic residue because otherwise, you cannot generate the whole mod p with the primitive root $a$. Thus, part (b) of your question follows from this idea.  
For part (a), 
$(\frac{-2}{p})=(\frac{p-2}{p})$ since  $-2 \equiv p-2$ (mod p) 
Therefore, you have to find primes p so that there exists $0<x<p$ such that 
$x^2 \equiv p-2$ (mod p) 
Alternatively, one can use Euler's Criterion: 
$a^{(p-1)/2} \equiv  (\frac{a}{p})$ (mod p) 
Solution:  Of course, maybe one can directly solve the equation above. Yet, I would do in the following way: 
$(\frac{-2}{p})=(\frac{2}{p})(\frac{-1}{p})= (\frac{2}{p})(−1)^{(p−1)/2}...(1)$ 
where the first equality comes from the multiplicative property of this operation. The second one is a consequence of Euler's Criterion (you can prove it): 
$(\frac{-1}{p})=(−1)^{(p−1)/2}$ 
Back to the equation: 
Case-1: If $p \equiv 3$ (mod 4) and 2 is a primitive root (mod p), then $(1)=1$. 
Case-2: If $p \equiv 1$ (mod 4) and 2 is a primitive root (mod p), then $(1)=-1$. 
In other cases, you can use the following proposition: 
The prime 2 is a quadratic residue modulo an odd prime p
if and only if $p \equiv ±1$ (mod 8). In other words: we have $(\frac{2}{p})=(−1)^{(p^2−1)/8}$. 
A: You probably have recently proved rules for evaluating $\left(\frac{-1}{p}\right)$ and $\left(\frac{2}{p}\right)$.  You can use these formulae  to answer part a.
$$\left(\frac{-2}{p}\right) =\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right).$$
For part b, suppose $a$ is a primitive root and so has order $p-1$.  Also suppose it is a quadratic residue, so that some $s^2 \equiv a \pmod{p}$.  What does the order of $s$ have to be and is this possible?
