About primitive root and quadratic residue.

Question

If p is an odd number and g is a primitive root (mod p), how can I prove that g is not a quadratic residue mod p?

Answer 1

Doing arithmetic modulo $\;p\;$ :

$$x^2=g\implies g^{\frac{p-1}2}=x^{p-1}=1\implies g\;\text{ has order less than}\;\;p-1$$

Answer 2

Given that $g$ is a primitive root $\bmod p$, then the smallest number $k>0$ for which $g^k \equiv 1 \bmod p$ is $k=\phi(p)$, and $g$ is coprime to $p$.

However, since $\phi(p)$ is even, if there is some $a$ such that $a^2 \equiv g \bmod p$, then we know that $a$ must also be coprime to $p$, giving us via Euler that $a^{\phi(p)}=1 \bmod p$ and so $g^{\phi(p)/2}=1 \bmod p$, which is not possible for a primitive root.