A Kakeya set in $\mathbb{F}_p^3$ of size $(2p^3+7p^2+2p-3)/8$ for $p \equiv 3 \pmod 4$

Let $p$ be an odd prime with $p \equiv 3 \pmod 4$. A Kakeya set in $\mathbb{F}_p^3$ is a set containing a full line in every direction: for each nonzero $v \in \mathbb{F}_p^3$ some translate $\{x + tv : t \in \mathbb{F}_p\}$ lies entirely inside it.

The assertion is that a Kakeya set of the following exact size exists:

The witness is an explicit algebraic construction. Writing $S \subseteq \mathbb{F}_p$ for the set of squares (including $0$), put

and take $K = K_1 \cup K_{2a} \cup K_{2b}$. The two quadratic conditions defining $K_1$ are what supply a line in every direction with $x$-component nonzero, while $K_{2a}$ and $K_{2b}$ cover the remaining directions lying in the plane $x = 0$.

The interest of the statement is quantitative rather than qualitative. Dvir's polynomial method gives a lower bound of order $p^3$ for every Kakeya set, so such sets necessarily occupy a positive proportion of $\mathbb{F}_p^3$; the open question is the optimal constant. This construction has density approaching $1/4$, and the exact lower-order terms recorded above are what make it a sharp benchmark rather than an asymptotic one.

Formalization note. The cardinality is stated in the cleared form $8|K| = \dots$ over $\mathbb{Z}$ so that no division appears and no rounding convention is implicit. The size has been confirmed by direct enumeration at $p = 3, 7, 11$, and the Kakeya property by exhaustive check over all nonzero directions at the smaller of those primes.