Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 11I (Number Theory)

A primitive root modulo $p$ is an integer $g$ whose residue class generates the multiplicative group of a finite field ${\mathbb{F}}_p^{\times }$. Equivalently, its multiplicative order modulo $p$ is $p-1$, so

For $k=0$, the convention $0^0=1$ gives $S_0=p\equiv 0(\mathrm{mod}p)$. Now suppose $1\le k<p-1$ and choose a primitive root $g$ modulo $p$. The nonzero residues are $g^j$ for $0\le j\le p-2$, while the $x=0$ term vanishes. HenceThis is a finite geometric series. Since $g^k\not\equiv 1(\mathrm{mod}p)$ but $g^{k(p-1)}\equiv 1(\mathrm{mod}p)$ by Fermat's little theorem, multiplication by the nonzero residue $g^k-1$ givesThe factor $g^k-1$ is invertible modulo the prime number $p$, so

By Fermat's little theorem,Thus the sum of $g$ over ${\mathbb{F}}_p^3$ counts, modulo $p$, the triples on which the ternary quadratic form $f$ vanishes.

Expand $f^{p-1}$ with the multinomial theorem. A term indexed by $r+s+t=p-1$ contributes a constant multiple ofAt least one of $r,s,t$ is strictly less than $(p-1)/2$. Its corresponding exponent is therefore in $\{0,1,\dots ,p-2\}$, so part (b), including its $k=0$ case, makes that power sum zero modulo $p$. Every term in the expansion consequently vanishes after summation. The constant term contributes $p^3\equiv 0$, and therefore

The zero triple is one solution of $f=0$. If it were the only solution, the displayed sum would be congruent to $1$ rather than $0$. Hence there is another triple $(x,y,z)$, not all divisible by $p$, satisfying