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


2 Answers 2


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):


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 $(\frac{2}{p})=(−1)^{(p^2−1)/8}$.


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?


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