Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 11I / c / Solution

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