Exercise 5.15

Proof. Let $g$ be a generator of ${𝔽}_p^{\text{∗}}$. Then $g$ has order $p-1$, thus $\rho =g^{(p-1)∕5}$ has order $5$.

Let

As $0={\rho }^5-1=(\rho -1)(1+\rho +{\rho }^2+{\rho }^3+{\rho }^4)$ and $\rho \ne 1$, then $1+\rho +{\rho }^2+{\rho }^3+{\rho }^4=0$, thus

This shows that $\alpha ,\beta$ are the roots in ${𝔽}_p$ of $x^2+x-1$, so that ${\alpha }^2+\alpha -1=0$.

Thus $4{\alpha }^2+4\alpha -4={(2\alpha +1)}^2-5=0$ : $\bar{5}$ is a square in ${𝔽}_p^{\text{∗}}$ and $\left(\frac{5}{p}\right)=1$. □