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If member sizes modulo a prime number $p$ avoid $L$ while distinct intersections lie in $L$, evaluation at characteristic vectors of sets gives a diagonal matrix with nonzero diagonal. After multilinear reduction on the Boolean cube, these polynomials remain linearly independent and have degree at most $|L|$. Counting square-free monomials proves $|\mathcal{F}|\le {\sum }_{j=0}^{|L|}\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. On a fixed uniform layer, the low-degree evaluation rank on a uniform layer sharpens this to $\left(\genfrac{}{}{0ex}{}{n}{|L|}\right)$ when $|L|$ does not exceed the layer rank.

Put $m=p^a$, with $p$ a prime number, and $s=m-1$. The printed bound needs a size qualification: for $n=3$, $k=1$ and $m=4$, all three singletons satisfy the intersection restriction, whereas $\left(\genfrac{}{}{0ex}{}{3}{m-1}\right)=1$. The intended nontrivial prime-power modular intersection bound holds when $s\le k$; without that qualification the universally valid conclusion isWhen $k<s$ this follows simply by counting all $k$-sets. We prove the sharper intended form for $s\le k$.

Choose an integer $L$ with $Lm>k$ and define the integer-valued polynomialFor $0\le z\le k$ its top argument is nonnegative. The bottom argument has all its first $a$ base-$p$ digits equal to $p-1$. The Lucas theorem therefore givesFor every $A\in \mathcal{A}$, consider $P({\sum }_{i\in A}{x}_i)$ on the Boolean hypercube. Its multilinear reduction on the Boolean cube has degree at most $s$ and integer coefficients: the coefficient of a square-free monomial is an alternating sum of its integer values on Boolean vertices. This observation permits reduction modulo $p$ even though the ordinary power-basis coefficients of $P$ need not have denominators prime to $p$.

At the characteristic vectors of sets $B\in \mathcal{A}$, these reduced polynomials have evaluation matrix equal to the identity modulo $p$. Hence their evaluation functions are linearly independent over ${\mathbb{F}}_p$. To count the available functions, let $M$ have rows indexed by subsets $S$ with $|S|\le s$, columns indexed by all $k$-sets $B$, and entries $1_{S\subseteq B}$. Over the rational numbers, for $|S|=j$,Thus the matrix rank over the rational numbers is at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Every larger integer minor is zero and remains zero modulo $p$, so the same rank bound holds over ${\mathbb{F}}_p$. This low-degree evaluation rank on a uniform layer proves

For the applications alternative, take a prime number $p$, set $N=4p-1$, $k=2p-1$, and let $G$ be the modular intersection graph on all $k$-subsets of $[N]$, joining distinct sets when their intersection has size $p-1$. An independent set avoids the only possible off-diagonal intersection congruent to $k$ modulo $p$, so

First embed $A$ as $v_A=1_A/\sqrt{2p}$ in Euclidean space. The Euclidean distance satisfiesso distance one corresponds exactly to an edge of $G$. Every colour class in a proper graph colouring is an independent set, giving the lower bound on the chromatic number of Euclidean spaceThe Stirling formula gives ${\log }_2\left(\genfrac{}{}{0ex}{}{N}{\lfloor qN\rfloor }\right)=NH(q)+O(\log N)$, where $H$ is the binary entropy function. Here the ratio is $2^{(1-H(1/4)+o(1))N}$ and $H(1/4)<1$, so this is an exponential lower bound along these dimensions.

Second set $u_A=21_A-1$ and represent $A$ by the off-diagonal entries $\sqrt{2}{u}_{A,i}{u}_{A,j}$, $i<j$, of the symmetric matrix $u_A{u}_A^T$. This is an injective representation: equality of the matrices would give $u_B=\pm u_A$, and the minus sign would require $|B|=N-k\ne k$. The common diagonal can be omitted without changing distances. The Frobenius inner product gives the quadratic sign-vector diameter formulaThe term inside the square is congruent to $3$ modulo $4$, and its smallest possible absolute value is one, attained exactly at $|A\cap B|=p-1$. Such pairs exist. Consequently the diameter pairs of this finite set in dimension $D=\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ are precisely the edges of $G$. Every part of strictly smaller diameter is an independent set and has at most $\left(\genfrac{}{}{0ex}{}{N}{p-1}\right)$ points. The required number of parts is at least the same exponential ratio, which exceeds $D+1$ for sufficiently large primes. Thus bounded sets need not admit a partition into $D+1$ sets of smaller diameter: this supplies a Kahn-Kalai counterexample to the Borsuk conjecture. Both applications use the prime case of the proved prime-power result.

The Frankl-Wilson theorem is a striking example of the polynomial method in combinatorics: an intersection restriction creates a family of linearly independent low-degree polynomials, whose dimension bounds the set family. Let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have $s$ elements, and let $\mathcal{F}$ be a set family such that member sizes modulo $p$ avoid $L$ while every distinct pair has intersection size in $L$ modulo $p$. Its general modular form givesIf all members have the same size $k$ and $s\le k$, the uniform refinement gives $|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$.

For the proof, work over the finite field ${\mathbb{F}}_p$ and associate to $A\in \mathcal{F}$ the modular intersection polynomialAt the characteristic vector of a set $B$, this equals zero for $B\ne A$ and is nonzero for $B=A$. A linear relation among the evaluation functions therefore has every coefficient zero. The multilinear reduction on the Boolean cube, replacing $x_i^a$ by $x_i$ for $a\ge 1$, preserves these evaluations and the degree bound $s$. There are ${\sum }_{j=0}^s\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ possible square-free monomials, proving the general form.

For uniform members, the low-degree evaluation rank on a uniform layer supplies the sharper dimension bound. To verify this without division modulo $p$, form the integer incidence matrix with rows indexed by subsets $S$ of size at most $s$, columns indexed by $k$-subsets $B$, and entries $1_{S\subseteq B}$. Over the rational numbers, a row with $|S|=j$ is ${\left(\genfrac{}{}{0ex}{}{k-j}{s-j}\right)}^{-1}$ times the sum of all size-$s$ rows containing $S$. Its matrix rank is at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Every larger integer minor is consequently zero and remains zero modulo $p$, giving the same bound over ${\mathbb{F}}_p$. The nonzero diagonal evaluation matrix of the $P_A$ then forces $|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$.

The theorem, proved by Frankl and Wilson in 1981, extends the theme of the Ray-Chaudhuri–Wilson theorem from ordinary intersection sizes to residues. Modular restrictions can be weaker than exact intersection restrictions yet still force unexpectedly small families. Here are three consequences from one explicit modular intersection graph.

Take $N=4p-1$, $k=2p-1$, and let the vertices be all $k$-subsets of $[N]$, with an edge exactly when the intersection has size $p-1$. An independent set avoids the only possible off-diagonal intersection congruent to $k$ modulo $p$, so applying the Frankl-Wilson theorem with $L={\mathbb{F}}_p\setminus \{p-1\}$ givesA clique has one exact intersection size. Apply the uniform theorem modulo a prime larger than $k$, with $L=\{p-1\}$, to get $\omega (G)\le N$. Since there are $v=\left(\genfrac{}{}{0ex}{}{N}{k}\right)$ vertices, this gives the explicit asymmetric Ramsey number bound $R(N+1,B+1)>v$. It illustrates how algebraic constructions produce graphs with simultaneously restricted cliques and independent sets.

For a geometric application, map each vertex to $x_A=1_A/\sqrt{2p}$ in ${\mathbb{R}}^N$. Then $|x_A-x_B{|}^2=(k-|A\cap B|)/p$, so edges have unit Euclidean distance. Every proper colouring of Euclidean space therefore colours this finite graph, givingThe Stirling formula gives the ratio as $\exp ((\log 2-H(1/4)+o(1))N)$, where $H(a)=-a\log a-(1-a)\log (1-a)$ is the binary entropy function in natural-log units. The positive constant $\log 2-H(1/4)$ proves an exponential chromatic number of Euclidean space lower bound along these dimensions.

Finally, a quadratic embedding converts this same forbidden intersection into a forbidden diameter pair. Set $u_A=21_A-1$ and $Q_A=u_A{u}_A^T$. The quadratic sign-vector diameter formula follows from the Frobenius inner product:For distinct members the smallest possible absolute inner product is one, attained exactly when the intersection is $p-1$. Thus maximum-distance pairs are exactly edges of $G$. The matrices have common diagonal; using their off-diagonal coordinates, multiplied by $\sqrt{2}$, realizes them isometrically in $D=\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ dimensions. They are distinct because equal outer products require $u_A=\pm u_B$, and complements have size $2p$, not $k$.

Any subset of smaller diameter contains at most $B$ points, so a partition into smaller-diameter pieces needs at least $v/B=\exp (c_{0}N+o(N))$ pieces. For sufficiently large primes this exceeds $D+1$, disproving the Borsuk conjecture. This is the mechanism behind the Kahn-Kalai counterexample to the Borsuk conjecture: the number of required pieces grows exponentially in the original set-system parameter while dimension grows quadratically. Their 1993 paper establishes the resulting high-dimensional counterexample.

Use the uniform Frankl-Wilson theorem: if $p$ is a prime number, $L\subseteq {\mathbb{F}}_p$ has $s$ distinct residues, $s\le \min \{r,n-r\}$, and an $r$-uniform set family $\mathcal{F}$ has distinct-member intersection sizes in $L$ modulo $p$ while $r\mathrm{mod}p\notin L$, thenHere is a polynomial method in combinatorics proof. For each $A\in \mathcal{F}$, form the intersection polynomialover ${\mathbb{F}}_p$, and use multilinear reduction on the Boolean cube so its degree is at most $s$. At the characteristic vector of $B\in \mathcal{F}$, this evaluates to zero for $A\ne B$ and is nonzero for $A=B$. Consequently these evaluation functions are linearly independent.

The low-degree evaluation rank on a uniform layer bounds their space by $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. To justify that step without division by a possibly zero residue, form the integer incidence matrix with rows $T\subseteq [n]$, $|T|\le s$, columns $r$-sets $B$, and entries $1_{T\subseteq B}$. Over the rational numbers, a row with $|T|=j$ equals ${\left(\genfrac{}{}{0ex}{}{r-j}{s-j}\right)}^{-1}$ times the sum of the rows of all $s$-sets containing $T$. Thus its rational matrix rank is at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Every larger minor is an integer determinant equal to zero and stays zero modulo $p$, so the same upper bound holds over ${\mathbb{F}}_p$. This proves the theorem.

For the forbidden midpoint problem, split $\mathcal{A}$ according to whether its members contain coordinate 1. Leave the containing half unchanged; complement every member of the avoiding half. Both resulting families consist of $2p$-sets containing 1, and complementation preserves their internal intersection sizes becauseWithin either transformed family, distinct intersections lie between 1 and $2p-1$ and are not $p$. Their residues are therefore in $L=\{1,\dots ,p-1\}$, whereas the set-size residue is $2p\equiv 0$. Apply the Frankl-Wilson theorem to each half and add:The factor 2 comes from splitting into two families. The unsplit family can contain disjoint sets, whose intersection residue zero equals the set-size residue and prevents this direct application. Removing the shared coordinate after splitting also gives the stronger complement splitting for forbidden midpoint intersections bound $2\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$.

For the construction, fix a $(p+1)$-set $B$ and let $R=[4p]\setminus B$, of size $3p-1$. The fixed-core construction avoiding midpoint intersections isTwo members of $\mathcal{F}$ intersect in at least $p+1$ points; the same is true for two complements. An intersection across the two halves has size $|X\setminus Y|\le p-1$. No intersection is $p$. The halves are disjoint because one contains all of $B$ and the other none, so

The uniform modular Frankl-Wilson theorem states the following. Let $p$ be prime, let $L\subseteq {\mathbb{F}}_p$ have size $s$, and let $\mathcal{F}$ be a rank-$k$ family on $[N]$, with $0\le s\le \min \{k,N-k\}$. Assume $k\mathrm{mod}p\notin L$ and $|A\cap B|\mathrm{mod}p\in L$ for every distinct $A,B\in \mathcal{F}$. ThenThe restrictions concern distinct pairs; the diagonal size residue is explicitly excluded from $L$.

For each member $A$, over ${\mathbb{F}}_p$ form the intersection polynomialAt the characteristic vector $1_B$, its value vanishes when $B\ne A$ and is nonzero when $B=A$. Thus these functions are linearly independent on the whole rank-$k$ layer: evaluate any linear relation at each family member in turn. Replacing every positive power of $z_i$ by $z_i$ preserves all Boolean evaluations, so the functions are restrictions of multilinear polynomials of degree at most $s$.

We need the sharp low-degree evaluation rank on a uniform layer, not the larger whole-cube dimension ${\sum }_{j\le s}\left(\genfrac{}{}{0ex}{}{N}{j}\right)$. Let $M$ be the integer incidence matrix with rows indexed by subsets $S$ of size at most $s$, columns by rank-$k$ sets $B$, and entries $1_{S\subseteq B}$. Over the rationals, if $|S|=j\le s$, thenThis coefficient counts the possible $T$ inside each containing column set and is positive because $s\le k$. Consequently the degree-$s$ rows span all rows over $\mathbb{Q}$, giving rank at most $\left(\genfrac{}{}{0ex}{}{N}{s}\right)$. Every larger square minor is therefore the integer zero. Reducing those minors modulo $p$ shows the rank over ${\mathbb{F}}_p$ is also at most $\left(\genfrac{}{}{0ex}{}{N}{s}\right)$. This transfer avoids dividing by a binomial coefficient that might vanish modulo $p$. The independent functions $f_A$ lie in this row space, so their number is at most its rank. This proves the theorem in full.

For the requested forbidden-intersection application, split $\mathcal{A}$ into sets containing coordinate one and sets avoiding it. In the first part, delete one. Distinct original sets have intersection size between one and $2p-1$, excluding $p$. Their reduced intersections therefore lie between zero and $2p-2$, excluding $p-1$, so their residues belong to $L=\{0,\dots ,p-2\}$. The reduced set size is $2p-1$, whose residue is $p-1\notin L$. Frankl-Wilson on the ground set of size $4p-1$ bounds this part by $\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$.

For the avoiding part, first complement every set. The complements contain one, retain size $2p$, and satisfyDelete one and apply the same argument. The complement splitting for forbidden midpoint intersections yields the stronger boundThe split is essential: a family may contain complementary pairs with intersection zero, preventing direct use of $L=\{1,\dots ,p-1\}$ on the original layer.

The Borsuk conjecture asserted that a bounded subset of ${\mathbb{R}}^d$ of positive diameter can be partitioned into $d+1$ subsets of strictly smaller diameter. Here is the Kahn-Kalai counterexample to the Borsuk conjecture, implemented by balanced cut vectors. For each $X\in [4p{]}^{(2p)}$, define $v_X\in {\mathbb{R}}^{\left(\genfrac{}{}{0ex}{}{4p}{2}\right)}$ whose coordinate indexed by an unordered pair $\{i,j\}$ is one when exactly one endpoint lies in $X$, and zero otherwise. Complementary sets give the same cut. Choose the representative with $1\in X$, obtaining exactlydistinct points: a cut determines its two parts, and fixing the side containing one removes its only ambiguity. Every cut has $(2p{)}^2=4p^2$ occupied coordinates, so all points lie in the affine hyperplane with coordinate sum $4p^2$. Its dimension isWe can identify this hyperplane isometrically with ${\mathbb{R}}^d$; no claim of full affine span is needed.

For two side sets with $t=|X\cap Y|$, their symmetric difference has $4p-2t$ points. The two cuts disagree exactly on pairs with one endpoint in that symmetric difference, giving the balanced cut-vector distance formulaThe diameter is $2p$, attained exactly when $t=p$; such pairs exist even among representatives containing one. Thus a smaller-diameter part contains no pair with side intersection $p$. Since all representative sets already contain one, deleting that common coordinate and applying the preceding Frankl-Wilson argument bounds each part by $\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$. There is no extra factor two at this step. This is the cut construction described in the original Kahn-Kalai paper; the distance and part-size calculations above give the required justification directly.

An explicit choice is $p=13$. The point set lies in dimensionThe exact counts areTheir ratio requires at least $\lceil M/B\rceil =1562$ smaller-diameter parts, exceeding $d+1=1326$. In particular,Rescaling the point set to diameter one changes none of these partition counts. Therefore the Borsuk counterexample in dimension 1325 is fully explicit. This is a concrete dimension, not a claim that it is the smallest possible counterexample dimension.

There is a genuine qualification missing from the printed bound. With $n=3,r=1,m=4$, the family of three singletons has distinct intersections of size zero, which is not congruent to $r$ modulo $m$. Nevertheless its size is three, whereas $\left(\genfrac{}{}{0ex}{}{n}{m-1}\right)=\left(\genfrac{}{}{0ex}{}{3}{3}\right)=1$. Thus that bound cannot hold without a restriction on the parameters. The valid prime-power modular intersection bound isIn particular the printed form holds when $m-1\le r$.

Put $m=p^a$ with $p$ a prime number, and $d=m-1$. If $r<d$, the corrected bound is simply the total number $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ of $r$-sets. Assume henceforth $d\le r$. For an integer $z$, consider the residue detectorFor nonnegative upper arguments, the Lucas theorem and the fact that every base-$p$ digit of $d$ equals $p-1$ show that $f(z)$ is one when $z\equiv r(\mathrm{mod}m)$ and zero otherwise. The assertion also holds for negative upper arguments, using integer-valued generalized binomial coefficients. Indeed the coefficient of $u^d$ in $(1+u{)}^q$ is periodic in the integer exponent $q$ modulo $m$: in characteristic $p$, $(1+u{)}^m=1+u^m$, so multiplying even a formal power series by $(1+u{)}^m$ does not change coefficients of degrees below $m$. Reduce any exponent modulo $m$ to obtain the stated detector.

For each $A\in \mathcal{F}$, apply that detector to $|A\cap B|$. Its evaluations on $B\in \mathcal{F}$ form the identity matrix, since the diagonal intersection is $r$ and every off-diagonal intersection has a different residue. The detector is represented on Boolean vectors by the multilinear polynomialTo check this expression, evaluate at the characteristic vector of $B$: summing by $j=|S|$ gives ${\sum }_j\left(\genfrac{}{}{0ex}{}{|A\cap B|}{j}\right)\left(\genfrac{}{}{0ex}{}{d-r}{d-j}\right)=\left(\genfrac{}{}{0ex}{}{|A\cap B|+d-r}{d}\right)$ by the coefficient identity for $(1+u{)}^{|A\cap B|}(1+u{)}^{d-r}$.

Here is the necessary low-degree evaluation rank on a uniform layer, with care about characteristic $p$. Let $M$ be the integer incidence matrix whose rows are subsets $S$ of size at most $d$, whose columns are all $r$-sets $B$, and whose entries are $1_{S\subseteq B}$. Over the rational numbers, for $|S|=j\le d$ its row satisfiesThus its rational matrix rank is at most $\left(\genfrac{}{}{0ex}{}{n}{d}\right)$. Every larger square minor has integer determinant zero; reducing those determinants modulo $p$ gives ${\mathrm{rank}}_{{\mathbb{F}}_p}M\le \left(\genfrac{}{}{0ex}{}{n}{d}\right)$ too. This argument does not divide by a potentially zero element of the finite field. The identity evaluation matrix $(f_A(1_B){)}_{A,B\in \mathcal{F}}$ factors through these incidence rows, so its rank $|\mathcal{F}|$ is at most $\left(\genfrac{}{}{0ex}{}{n}{d}\right)$. This proves the qualified bound.

Choose the two-applications alternative. First, the polynomial method in combinatorics yields a constructive lower bound for a diagonal Ramsey number. Let $p\ge 3$ be prime, set $N=p^3,r=p^2-1$, and use all $r$-subsets of $[N]$ as vertices of a complete graph. Color a pair red when its intersection size is $-1$ modulo $p$, and blue otherwise. A blue clique has size at most $D=\left(\genfrac{}{}{0ex}{}{N}{p-1}\right)$ by the proved bound with modulus $p$. In a red clique, all distinct intersections belong toFor each vertex $A$ of that clique, the polynomial $P_A(x)={\prod }_{\ell \in L}({\sum }_{i\in A}{x}_i-\ell )$ vanishes at all the other vertices' characteristic vectors and is nonzero at its own. Replace repeated powers $x_i^j$ by $x_i$ on Boolean vectors. Its degree is at most $p-1$, so the rational version of the same incidence-rank argument bounds a red clique by $D$ as well. Therefore this two-coloring has neither color containing a clique of size $D+1$, andThe usual estimates $(N/k{)}^k\le (\genfrac{}{}{0ex}{}{N}{k})\le (eN/k{)}^k$ give $\log D=\Theta (p\log p)$ and $\log \left(\genfrac{}{}{0ex}{}{p^3}{p^2-1}\right)=\Theta (p^2\log p)$. Thus along this explicit sequence the diagonal Ramsey number exceeds $\exp (c(\log s{)}^2/\log \log s)$ for some absolute $c>0$. This is substantially larger than any fixed power of $s$.

Second, obtain an exponential lower bound for the chromatic number of Euclidean space. Let $N=4p-1,r=2p-1$, and join two $r$-subsets exactly when their intersection has size $p-1$. An independent set in this graph avoids that intersection. Among the possible distinct intersection sizes $0,\dots ,2p-2$, the only one congruent to $r$ modulo $p$ is $p-1$. Hence every independent set has size at most $\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$. Embed each vertex as $v_A=1_A/\sqrt{2p}$ in Euclidean space. Thenso the graph's edges have exactly Euclidean distance one. Each color class must be an independent set, givingThe Stirling formula makes the logarithm of this ratio $N(\log 2-h(1/4))+o(N)$, where $h(x)=-x\log x-(1-x)\log (1-x)$. Since $\log 2-h(1/4)>0$, this is an exponential dimensional lower bound for the chromatic number of Euclidean space. Both applications lie in the valid parameter range of the proved modular bound.

The Frankl-Wilson theorem in its uniform form says: let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have size $s\le \min (k,n-k)$, and let $\mathcal{F}\subseteq \left(\genfrac{}{}{0ex}{}{[n]}{k}\right)$ have $k\mathrm{mod}p\notin L$, while $|A\cap B|\mathrm{mod}p\in L$ for every distinct pair of members. Then

The modular-intersection, nonuniform form of the Frankl-Wilson theorem is as follows. Let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have $s$ elements, and let $\mathcal{F}\subseteq \mathcal{P}([n])$ satisfy $|A|\mathrm{mod}p\notin L$ for each $A\in \mathcal{F}$, whereas $|A\cap B|\mathrm{mod}p\in L$ for distinct $A,B\in \mathcal{F}$. ThenThis nonuniform Frankl-Wilson theorem allows different sizes and different diagonal residues; in particular it applies when all sizes have one common residue outside $L$.

To prove it by the polynomial method in combinatorics, work over the finite field ${\mathbb{F}}_p$ and associate to each member the intersection polynomialReplace each positive power of $x_i$ in its expansion by $x_i$ to obtain its multilinear reduction on the Boolean cube ${\tilde{P}}_A$. This preserves evaluation on every characteristic vector, whose coordinates are zero or one, and leaves degree at most $s$. At the characteristic vector $1_B$,It is zero for distinct $A,B$, and nonzero for $B=A$ by the diagonal hypothesis. If ${\sum }_{A\in \mathcal{F}}{c}_A{\tilde{P}}_A=0$, evaluating at each $1_B$ forces $c_B=0$. Thus the multilinear polynomials are linearly independent. They belong to the vector space with basis the square-free monomials ${\prod }_{i\in S}{x}_i$ for $|S|\le s$, whose dimension is the displayed sum. This proves the nonuniform form, including $s=0$, where distinct members are impossible and the empty product is one.

For the sharp uniform form, we need the low-degree evaluation rank on a uniform layer. Form the integer matrix $M$ whose rows are indexed by $S\subseteq [n]$ with $|S|\le s$, columns by $k$-subsets $T$, and entries $M_{S,T}=1_{S\subseteq T}$. These are evaluations of the square-free monomials. Over the rational numbers, the row of a $j$-subset $S$ satisfiesIndeed, when $S\subseteq T$ the left side counts the $s$-subsets between $S$ and $T$, and otherwise both sides vanish. Since $s\le k$, the coefficient is a positive integer. Thus the degree-$s$ rows span all the rows over the rational numbers, giving ${\mathrm{rank}}_{\mathbb{Q}}M\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Every square submatrix of larger size has integer determinant zero, so its determinant remains zero modulo $p$. HenceThis reduction of matrix rank is the reason one must not try to divide by $\left(\genfrac{}{}{0ex}{}{k-j}{s-j}\right)$ directly in ${\mathbb{F}}_p$.

The evaluations of the ${\tilde{P}}_A$ on the $k$th uniform layer of the Boolean cube lie in the row linear span of $M$ over ${\mathbb{F}}_p$. They are linearly independent as functions there, because restriction to the columns corresponding to $\mathcal{F}$ still gives the nonzero diagonal evaluation matrix. Their number is consequently at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. This completes the proof of the uniform Frankl-Wilson theorem as well.

For the odd-cardinality problem the sharper Oddtown theorem follows directly from linear algebra. Let $v_A=1_A\in {\mathbb{F}}_2^n$ be the characteristic vector. The parity conditions sayFor any linear combination ${\sum }_A{c}_A{v}_A=0$, take its dot product with $v_B$. All off-diagonal terms vanish, leaving $c_B=0$. Hence these vectors are linearly independent in an $n$-dimensional vector space, givingThe singleton sets show that the bound can be attained. Notice why the constant term in the general nonuniform Frankl-Wilson theorem is unnecessary here: these dot products already give a diagonal evaluation matrix using only the $n$ coordinate functions.

For even $n$, partition $[n]$ into the pairs $\{1,2\},\{3,4\},\dots ,\{n-1,n\}$ and take every union of a selection of these pairs. This gives exactly $2^{n/2}$ distinct sets, including the empty set. Each has even size; every pairwise intersection is again a union of entire pairs and therefore has even size. This attains the proposed bound.

For any set family satisfying the even-cardinality conditions, let $W$ be the linear span of its characteristic vectors over ${\mathbb{F}}_2$. Now the self-dot products vanish as well as the off-diagonal ones. By bilinearity, $u\cdot v=0$ for all $u,v\in W$, so $W\subseteq W^{\perp }$: it is a self-orthogonal binary subspace.

For completeness the standard bilinear form on ${\mathbb{F}}_2^n$ is nondegenerate, because a vector orthogonal to every coordinate vector has every coordinate zero. Thus the map $v\mapsto (w\mapsto v\cdot w)$ from ${\mathbb{F}}_2^n$ onto the dual space $W^{\ast }$ is surjective: every linear functional on $W$ extends to the whole vector space by extending a basis, and every such functional on ${\mathbb{F}}_2^n$ is a dot product with a vector. Its kernel is $W^{\perp }$. The rank-nullity theorem consequently gives $\dim W^{\perp }=n-\dim W$. The inclusion $W\subseteq W^{\perp }$ implies $\dim W\le n/2$. Finally, distinct members have distinct characteristic vectors, all in $W$, and a $k$-dimensional vector space over ${\mathbb{F}}_2$ has exactly $2^k$ elements. ThereforeThis proves the Eventown theorem and also explains the suggested dimension contradiction if the set family had more than $2^{n/2}$ members.

Let $m=p^a$ and let an $r$-uniform set family have no distinct pair whose intersection is congruent to $r$ modulo $m$. Its size is bounded as displayed; the customary nontrivial form $\left(\genfrac{}{}{0ex}{}{n}{m-1}\right)$ requires $m-1\le r$. The Lucas theorem makes $\left(\genfrac{}{}{0ex}{}{z-r+m-1}{m-1}\right)$ a residue detector modulo $p$, equal to one exactly when $z\equiv r(\mathrm{mod}m)$. Its Boolean evaluation functions have degree at most $m-1$ and a diagonal evaluation matrix on the family. The low-degree evaluation rank on a uniform layer supplies the sharp dimension bound. Without the size qualification, three singletons with $n=3,r=1,m=4$ contradict the customary printed form, since $3>\left(\genfrac{}{}{0ex}{}{3}{3}\right)$.

If a $k$-uniform set family on an $n$-element ground set has at most $s$ possible intersection sizes between distinct members, then its size is at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. The proof uses the linear independence of incidence polynomials on the $k$-slice.