1. ## Help

We will show that $\displaystyle 3$ is a primitive root modulo $\displaystyle p$. If $\displaystyle 3$ is not a primitive root then $\displaystyle 3$ must be a quadradic residue. This is because $\displaystyle 3^{(p-1)/2}\equiv \pm 1(\bmod p)$. It cannot be $\displaystyle -1$ for that would force $\displaystyle p-1$ to be the order of $\displaystyle 3$ since $\displaystyle p-1$ is a power of $\displaystyle 2$. Therefore, there is $\displaystyle a$ such that $\displaystyle a^2 \equiv - 3(\bmod p)$. Let $\displaystyle x$ be the solution to $\displaystyle 2x\equiv a - 1(\bmod p)$. Then, $\displaystyle (2x)^3 \equiv (a-1)^3 (\bmod p)$ and so $\displaystyle 8x^3 \equiv (a^3+3a) - (1+3a^2)\equiv 8(\bmod p)$. Thus, we see that $\displaystyle x$ has order $\displaystyle 3$. That gives us a contradiction because then $\displaystyle 3$ divides $\displaystyle (p-1)$ which is impossible. Since $\displaystyle 3$ is a primitive root it means $\displaystyle (3/p)=-1$ and so $\displaystyle 3^{(p-1)/2}\equiv -1(\bmod p)$. We see that $\displaystyle p$ divides $\displaystyle 3^{(p-1)/2}+1$.