= Solution
A <finite-field Kakeya set>, also called a finite-field Besicovitch set, is a subset containing a complete <affine line in a vector space> $a+\mathbb F_pv$ in every nonzero direction $v$, with scalar multiples counted as the same direction. The line's translate may depend on the direction.
For odd $p$, take the union of the lines $y=mx-m^2/4$, $m\in\mathbb F_p$. Completing the square gives $x^2-y=(m/2-x)^2$. Thus that union is exactly $K_0=\{(x,y):x^2-y\text{ is a square, including zero}\}$. For each fixed $x$ there are $(p+1)/2$ such $y$, so $|K_0|=p(p+1)/2$. It contains every nonvertical slope. Add the vertical line $x=0$, whose overlap with $K_0$ has $(p+1)/2$ points. The <tangent-line finite-field Kakeya construction> consequently has
$$
\boxed{|K|=\frac{p^2+2p-1}{2}=\frac12p^2(1+o(1)).}
$$
The prime two is irrelevant to this asymptotic statement and can be handled by taking its whole plane.
For the high-dimensional lower bound, we give a direct <polynomial method in combinatorics> proof, which yields a stronger estimate. Suppose a finite-field Kakeya set $K\subseteq\mathbb F_p^n$ has fewer than $\binom{p+n-1}{n}$ points. That binomial coefficient is the <dimension of a bounded-total-degree polynomial space> of <polynomials> of total degree at most $p-1$. Evaluation at the points of $K$ imposes fewer linear conditions than this dimension. By the <rank-nullity theorem>, there is a nonzero <polynomial> $P$ of total degree $d\leq p-1$ vanishing on $K$.
For every nonzero $v$ choose the line $a_v+tv\subseteq K$. The <polynomial> $P(a_v+tv)$ in $t$ has degree at most $d<p$ and vanishes at all $p$ field elements, so it vanishes identically. Its coefficient of $t^d$ is the leading <homogeneous polynomial> part $P_d(v)$. Thus $P_d(v)=0$ for every nonzero $v$. The case $d=0$ is already impossible for a nonzero constant vanishing on a nonempty set; for $d>0$, the absence of a constant term in this <homogeneous polynomial> also gives $P_d(0)=0$.
A <polynomial> of degree less than $p$ in each variable cannot vanish at every point of $\mathbb F_p^n$ unless it is zero. To see this, induct on $n$: fix the first $n-1$ variables, apply the one-variable root bound in the last variable, then apply the induction hypothesis to each resulting coefficient <polynomial>. This forces $P_d=0$, contradicting the choice of the leading part. We have proved the <finite-field Kakeya polynomial bound>
$$
|K|\geq\binom{p+n-1}{n}\geq\frac{p^n}{n!}.
$$
At $n=15$ this implies the required bound with an explicit absolute constant:
$$
\boxed{|K|\geq\frac{p^{15}}{15!}\geq\frac{p^9}{15!},\qquad c=1/15!.}
$$
The argument is valid in every prime characteristic and does not invoke the lower bound as an unproved theorem.
Back to article page
