Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 109 / 2 / b

We use the coefficient form of the Combinatorial Nullstellensatz: if a polynomial of degree ${\sum }_i{t}_i$ has nonzero coefficient at ${\prod }_i{x}_i^{t_i}$, then it cannot vanish on every product set ${\prod }_i{S}_i$ with $|S_i|>t_i$.

Write the terms as $(a_i,b_i)$ and, over ${\mathbb{F}}_p$, putandConsiderA direct multinomial-coefficient calculation, using Wilson's theorem, shows that the coefficient of ${\prod }_{i=1}^{4p-2}{x}_i$ in $F$ is nonzero. The Combinatorial Nullstellensatz with every $S_i=\{0,1\}$ therefore gives an indicator vector $x$ for which $F(x)\ne 0$.

The zero indicator does not work. The first two factors force the selected vectors to have both coordinate sums zero. If the support size is not divisible by $p$, Fermat's little theorem makes $1-S^{p-1}=0$, while the final product also vanishes; hence the support has size $p$, $2p$, or $3p$. Size $p$ proves the claim, and size $3p$ reduces to part (a). For size $2p$,so the bracket vanishes, again contradicting $F(x)\ne 0$. Only the size-$p$ and reducible size-$3p$ cases remain, and either yields the required zero-sum $p$-subsequence.