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Kahn and Kalai represented uniform subsets by quadratic sign vectors so that diameter pairs correspond to one forbidden intersection size. The Frankl-Wilson theorem bounds each smaller-diameter part, forcing exponentially many parts while the ambient dimension grows only quadratically.

Fix a prime number $p>r$. Distinct members of an intersecting $r$-uniform family have intersection size inwhereas every member has size $r\notin L$ modulo $p$. The Frankl-Wilson theorem with $|L|=r-1$ therefore givesFor fixed $r$,This is the asserted asymptotic weakening of the Erdős-Ko-Rado theorem.

Solved by gpt-5.6-sol high.

Partition the family into complementary pairs $\{A,[4p]\setminus A\}$. Choose at most one member from each pair to obtain ${\mathcal{A}}^{\prime }$ with $|{\mathcal{A}}^{\prime }|\ge |\mathcal{A}|/2$. Distinct members of ${\mathcal{A}}^{\prime }$ are not disjoint, and the hypothesis excludes intersection size $p$. Since their intersection sizes lie between $1$ and $2p-1$, they therefore lie modulo $p$ inEvery member has size $2p\equiv 0(\mathrm{mod}p)$, which is outside $L$. The Frankl-Wilson theorem givesand hence

Solved by gpt-5.6-sol high.

One uniform form of the Frankl-Wilson theorem is as follows. Let $p$ be prime and let $L\subseteq {\mathbb{F}}_p$ have $s\le \min \{k,n-k\}$ elements. If $\mathcal{A}\subseteq [n{]}^{(k)}$ satisfiesthenThe proof assigns to each set a degree-$s$ polynomial that vanishes on the incidence vectors of all other members but not on its own. These functions are linearly independent in the space spanned by square-free monomials of degree $s$, yielding the dimension bound.

Solved by gpt-5.6-sol high.

The required prime-power form of the Frankl-Wilson theorem is: if $A_1,\dots ,A_m\subseteq [n]$, no $|A_i|$ is divisible by $p^r$, and every intersection of $k$ distinct members has size divisible by $p^r$, thenIt follows by associating to each $A_i$ its incidence vector augmented by a constant coordinate and applying the Frankl-Wilson polynomial independence lemma to the $(k-1)$ layers of multilinear intersection polynomials. The hypotheses make the diagonal evaluations nonzero modulo $p^r$ and every $k$-fold off-diagonal evaluation zero; independence leaves at most $(k-1)(n+1)$ polynomials. Applying the theorem gives the desired bound.

When $r=1$, the argument works directly over ${\mathbb{F}}_p$ without the constant-coordinate lift and yields the stronger boundTherefore a family of size $(k-1)(n+1)$ does not exist.

Solved by gpt-5.6-sol high.