{c}

There is a constant $C<3$ such that every <cap set> in $\mathbb F_3^n$ has cardinality less than $C^n$.

For the <polynomial method in combinatorics>, use
$$
T(x,y,z)=\prod_{i=1}^n\bigl(1-(x_i+y_i+z_i)^2\bigr).
$$
Over $\mathbb F_3$, this is the <indicator function> of $x+y+z=0$. On $A^3$ it is therefore a diagonal tensor with $|A|$ nonzero diagonal entries. Every <monomial> in its expansion has individual exponents at most two and total degree at most $2n$, so one of its three variable blocks has degree at most $2n/3$. Grouping terms according to such a block and using the <slice rank of a diagonal tensor> gives
$$
|A|=\operatorname{slice\ rank}(T|_{A^3})\leq3m_n,
$$
where $m_n$ is the number of $\alpha\in\{0,1,2\}^n$ with $\sum_i\alpha_i\leq2n/3$. If $X_i=1-\alpha_i$ are <independent random variables> uniform on $\{-1,0,1\}$, then
$$
\frac{m_n}{3^n}=\mathbb P\left(\sum_iX_i\geq\frac n3\right).
$$
Any exponential upper bound for this <tail probability> gives $m_n\leq(3-\epsilon)^n$ and hence the result after absorbing the factor three and finitely many small dimensions into $C<3$.

Ellenberg–Gijswijt cap-set bound

Assume first that $A\subseteq\mathbb F_3^n$ is a <cap set>. Over the <finite field> $\mathbb F_3$, define
$$
T(x,y,z)=\prod_{i=1}^n\left(1-(x_i+y_i+z_i)^2\right).
$$
Since $1-t^2$ is one at $t=0$ and zero at $t=\pm1$, $T$ is the <indicator function> of $x+y+z=0$. If $x,y,z\in A$ satisfy this equation, then either they are all equal or they are three distinct points; the latter is excluded. Thus $T|_{A^3}$ is a <diagonal tensor> with every diagonal entry equal to one, and the <slice rank of a diagonal tensor> gives
$$
\operatorname{slice\ rank}(T|_{A^3})=|A|.
$$

Expand $T$ as a <polynomial>. Every variable has exponent at most two, and every <monomial> has total degree at most $2n$. Splitting a monomial's degree among its $x$-, $y$-, and $z$-blocks, at least one block has degree at most $2n/3$. Assign each monomial to one such block and group together terms with the same low-degree block monomial. Each group is one slice, so
$$
|A|\leq3m_n,
$$
where
$$
m_n=\left|\left\{\alpha\in\{0,1,2\}^n:\sum_i\alpha_i\leq\frac{2n}{3}\right\}\right|.
$$
If $X_i=1-\alpha_i$, then the $X_i$ are <independent random variables> uniformly distributed on $\{-1,0,1\}$ and
$$
\frac{m_n}{3^n}
=\mathbb P\left(\sum_iX_i\geq\frac n3\right)
\leq e^{-n/12}
$$
by the given <tail probability> bound. Consequently every cap set satisfies
$$
|A|\leq3^{n+1}e^{-n/12}.
$$

Put $C_0=3e^{-1/24}<3$. For $n\geq27$, the preceding bound is at most $C_0^n$. For each of the finitely many $1\leq n<27$, a cap set is a proper subset of $\mathbb F_3^n$, so its size is at most $3^n-1$. We may therefore choose
$$
\max\left\{C_0,\max_{1\leq n<27}(3^n-1)^{1/n}\right\}<C<3.
$$
Every cap set then has size strictly below $C^n$. Equivalently,
$$
\boxed{|A|\geq C^n\quad\Longrightarrow\quad
\text{$A$ contains distinct $x,y,z$ with $x+y+z=0$.}}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 147 / 4 / Solution

If
$$
T(x,y,z)=\begin{cases}c_x&x=y=z,\\0&\text{otherwise},\end{cases}
$$
on a finite set $X$, and every $c_x$ is nonzero, then the <slice rank> of $T$ is $|X|$. The upper bound uses one $x$-slice for each diagonal entry. For the lower bound, restrict any shorter slice decomposition to a common kernel of the coefficient functions in two slice directions; the remaining diagonal matrix has rank larger than the number of slices available in the third direction, a contradiction.

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