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For a finite-field Kakeya set $A\subseteq {\mathbb{F}}_q^n$, a smaller cardinality would give a nonzero polynomial of total degree of a polynomial at most $q-1$ vanishing on $A$, by the dimension of a bounded-total-degree polynomial space. Its restriction to each complete affine line in a vector space has $q$ roots of a polynomial and degree less than $q$, hence vanishes identically. The top homogeneous polynomial part then vanishes in every direction. A polynomial of degree less than $q$ in each variable cannot vanish everywhere on ${\mathbb{F}}_q^n$ unless it is zero, giving a contradiction. For the prime field, this is the polynomial nonvanishing below the field size.

A subset of ${\mathbb{F}}_q^n$ is a finite-field Kakeya set when it contains an affine line in a vector space in every one-dimensional direction. The line position may depend on the direction. The polynomial method in combinatorics yields the finite-field Kakeya polynomial bound on its cardinality.

Subtracting the first equation from the third gives $x-z=3/2$. The second equation then becomes $(\alpha -2)x=\beta -3$, and the third fixes $y=(7-7x)/5$. The linear system has the following three consistency cases:The first case is a unique solution, the second is an affine line in a vector space of solutions, and the third is an inconsistent linear system.

A finite-field Kakeya set, also called a finite-field Besicovitch set, is a subset containing a complete affine line in a vector space $a+{\mathbb{F}}_{p}v$ 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 hasThe 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 $\left(\genfrac{}{}{0ex}{}{p+n-1}{n}\right)$ 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 $d\le p-1$ 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 boundAt $n=15$ this implies the required bound with an explicit absolute constant:The argument is valid in every prime characteristic and does not invoke the lower bound as an unproved theorem.

Apply the polynomial method in combinatorics to the finite-field Kakeya set $A$. We will actually obtain the stronger estimateLet $\mathcal{P}$ be the vector space over the prime field ${\mathbb{F}}_p$ of multivariate polynomials in $n$ variables of total degree of a polynomial at most $p-1$. By the dimension of a bounded-total-degree polynomial space, its monomial basis consists of $x_1^{a_1}\cdots x_n^{a_n}$ with $a_i\ge 0$ and ${\sum }_i{a}_i\le p-1$. Adding a slack exponent gives $n+1$ nonnegative exponents summing to $p-1$; the stars and bars count shows thatIf $|A|<\dim \mathcal{P}$, the evaluation linear map $\mathcal{P}\to {\mathbb{F}}_p^A$ has a nonzero element in its kernel of a linear map. Thus there is a nonzero polynomial $P$ of total degree of a polynomial $d\le p-1$ vanishing at every point of $A$. Its degree cannot be zero, since $A$ contains a line and is nonempty. Write $P_d$ for its nonzero top homogeneous polynomial part.

For every nonzero $v\in {\mathbb{F}}_p^n$, the directional hypothesis supplies an affine line in a vector spaceDirections are one-dimensional vector subspaces; rescaling a representative does not change this line. The univariate polynomial $P(a_v+tv)$ has degree at most $d<p$ and vanishes for all $p$ values of $t$. The root bound for a polynomial makes it the zero polynomial. Its coefficient of $t^d$ is exactly $P_d(v)$, so $P_d(v)=0$ for every nonzero $v$. Because $d>0$, $P_d(0)=0$ too.

To finish, we prove the relevant polynomial nonvanishing below the field size. A polynomial over ${\mathbb{F}}_p$ of degree at most $p-1$ in each variable cannot vanish on all of ${\mathbb{F}}_p^n$ unless it is zero. Induct on $n$. The one-variable case is the root bound for a polynomial. For more variables, writeFixing the first $n-1$ coordinates gives a univariate polynomial with $p$ roots of a polynomial, so all its coefficients are zero. Hence every $Q_j$ vanishes on ${\mathbb{F}}_p^{n-1}$, and the induction hypothesis makes every $Q_j$ zero. Applying this to $P_d$, whose total degree of a polynomial is less than $p$, contradicts its choice as nonzero.

Therefore $|A|\ge \dim \mathcal{P}$. Finally,which proves the requested lower bound. The crucial observation in the finite-field Kakeya polynomial bound is that complete lines force the highest homogeneous polynomial part to vanish in every direction.

Writing the unknown vector as $(x,y,z{)}^T$, the first two linear equations give $x=z-1$ and $y=z-\beta$. Substitution into the third givesThus, when $\alpha \ne 2$, put $t=-\beta /(\alpha -2)$. There is exactly one solution:When $\alpha =2$, the remaining equation is $0=-\beta$. HenceThe second case is an affine line in a vector space, so it contains infinitely many solutions. These cases exhaust all real numbers $\alpha ,\beta$.

Subtract the two position vectors to obtain the direction $\mathbf{b}-\mathbf{a}=(-3,-1,3)$. The affine line in a vector space through the points is thereforeThe values $t=0,1$ recover the two specified points.

Expand the right-hand side in the orthonormal basis from part (b). Its coefficients areThus $\mathbf{b}=2\sqrt{2}{\mathbf{e}}_4+\sqrt{3}{\mathbf{e}}_6$. If $\mathbf{x}=c_0{\mathbf{e}}_0+c_4{\mathbf{e}}_4+c_6{\mathbf{e}}_6$, then matrix multiplication gives $A\mathbf{x}=4c_4{\mathbf{e}}_4+6c_6{\mathbf{e}}_6$. Matching coefficients fixes $c_4=\sqrt{2}/2$ and $c_6=\sqrt{3}/6$, while $c_0$ is arbitrary. Consequently all solutions areGeometrically, the solution set is an affine line in a vector space parallel to the kernel of $A$. The zero eigenvalue makes the component in that direction invisible to $A$; the other two components are uniquely determined. The displayed particular solution is perpendicular to the kernel, so it is also the point of this line closest to the origin.

Suppose every point of ${\mathbb{F}}_q^n$ lies on an affine line in a vector space meeting a subset $N$ in at least $m$ points, with $1\le m\le q$. A smaller $N$ would admit a nonzero multivariate polynomial of total degree at most $m-1$ vanishing on it, by the dimension of a bounded-total-degree polynomial space. Each selected line then has more roots of a polynomial than the degree of its restriction, so the polynomial vanishes at every point. The Schwartz-Zippel lemma forbids this because $m-1<q$. Thus the displayed bound holds, and is at least $m^n/n!$. This is a point-covering condition; the finite-field Kakeya set condition instead quantifies over directions.