Past exam of the options II course of the University of Oxford / 2004 / C / C2 / c / ii

Let $p=F_m$ be prime with $m\ge 1$. The preceding parts give $(3/p)=-1$, so Euler's criterion implies $3^{(p-1)/2}\equiv -1(\mathrm{mod}p)$. The multiplicative order $d$ of three divides $p-1=2^{2^m}$. Every proper divisor of this power of two divides $(p-1)/2$. If $d$ were proper, that half-power would be one, a contradiction. Thereforewhich means precisely that three is a primitive root modulo $p$.