Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 1I / Solution

Euler's criterion states that for an odd prime number $p$ and $p\nmid a$,where the right-hand side is the Legendre symbol.

Let $g$ be a primitive root modulo $p$. Its multiplicative order is $p-1$, so $g^{(p-1)/2}\ne 1$. This power has square $g^{p-1}=1$, and the only roots of $X^2-1$ modulo the odd prime $p$ are $\pm 1$. ThereforeEuler's criterion now gives $(g/p)=-1$, so every primitive root is a quadratic nonresidue.

Now let $p=2^{2^k}+1$ be a Fermat prime, with $k\ge 1$. The number of primitive roots modulo $p$ isThere are also exactly $(p-1)/2$ quadratic nonresidues. Since every primitive root is a nonresidue, these equally large sets coincide. Thus every quadratic nonresidue modulo $p$ is a primitive root, as summarized by Quadratic nonresidues modulo a Fermat prime.

Finally, $p\equiv 1(\mathrm{mod}4)$ and $p\equiv 2(\mathrm{mod}3)$. Quadratic reciprocity therefore givesHence $3$ is a quadratic nonresidue modulo $p$, and consequently

Solved by gpt-5.6-sol high.