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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 diagonal diagonal Ramsey number $R(s)$ is the least $n$ such that every red-blue edge coloring of $K_n$ contains a monochromatic $K_s$. More generally let $R(s,t)$ require a red $K_s$ or blue $K_t$. The base cases are $R(1,t)=R(s,1)=1$. Given $R(s-1,t)+R(s,t-1)$ vertices, select one vertex. Either it has at least $R(s-1,t)$ red neighbors or at least $R(s,t-1)$ blue neighbors. Applying the corresponding smaller problem and, when needed, adjoining that selected vertex givesInduction and Pascal's identity now prove existence and

The graph Ramsey number $R(s,t)$ is the least $N$ for which every graph on $N$ vertices has a clique of size $s$ or an independent set of size $t$; $R(s)=R(s,s)$ is the diagonal Ramsey number. The base cases are $R(2,t)=t$ and $R(s,2)=s$.

Assume the smaller parameters already have finite graph Ramsey numbers, and take $N=R(s-1,t)+R(s,t-1)$. For any vertex, either its neighbourhood has at least $R(s-1,t)$ vertices, or its nonneighbourhood has at least $R(s,t-1)$ vertices. In the first case it contains an independent $t$-set, or an $(s-1)$-clique which extends with the chosen vertex. In the second it contains an $s$-clique, or a $(t-1)$-independent set which extends with that vertex. Induction proves existence and

The diagonal Ramsey number $R(s)$ is the least positive integer $N$ such that every red-blue edge colouring of the complete graph $K_N$ contains a monochromatic copy of $K_s$.

For the existence and bound, write $R(r,s)$ for the least $N$ forcing either a red $K_r$ or a blue $K_s$. Given a vertex $v$ of $K_{R(r-1,s)+R(r,s-1)}$, the pigeonhole principle says that at least $R(r-1,s)$ incident edges are red or at least $R(r,s-1)$ are blue. In the first case those red neighbours contain a red $K_{r-1}$, which joins to $v$ to make a red $K_r$, or a blue $K_s$; the second case is symmetric. HenceStarting from $R(1,s)=R(r,1)=1$ and using Pascal's identity gives the binomial upper bound for a Ramsey numberThereforewhich proves both existence and the required estimate.

To determine $R(3)$, color a cycle graph $C_5$ red and its complement graph, also a $C_5$, yellow. This colors $K_5$ without a monochromatic triangle, so $R(3)>5$. At any vertex of a red-yellow $K_6$, at least three of the five incident edges have one color. If an edge among their three other endpoints has that color, it completes a monochromatic triangle with the chosen vertex; if none does, those three endpoints themselves form a triangle of the other color. Thus $R(3)\le 6$, and

The same local argument proves uniqueness on $K_5$. No vertex can have three incident edges of one color: their other endpoints would either contain an edge of that color or form a triangle of the other color. Every vertex therefore has vertex degree at most two in each color, and because the two degrees sum to four, each is exactly two. Each color class is consequently a regular graph of degree two on five vertices, hence a single $C_5$. Up to relabelling, the coloring is exactly the cycle-and-complement coloring above.

Finally, the monochromatic-triangle counting formula gives, for a coloring of $K_6$ with red degrees $d_v$,The bound is attained by coloring all edges of $K_{3,3}$ red and the two complementary copies of $K_3$ yellow: there are no red triangles and exactly two yellow triangles. The least requested number is therefore

The diagonal Ramsey number $R(k)$ is the least $n$ such that every red-blue coloring of $K_n$ contains a monochromatic $K_k$. The standard recursiongivesThus the converted $k^4$ is a superscript-order error; that bound is false for large $k$.

In a red-blue coloring, if the red spanning graph is connected it has a red spanning tree. Otherwise its red components are joined pairwise by blue edges, making the blue graph connected, so it has a blue spanning tree.

The result fails for three colors. Color the six edges of $K_4$ by its three perfect matchings, one color per matching. Every monochromatic graph then consists of two disjoint edges and has no spanning tree.

The diagonal Ramsey number $R(t)$ is the least positive integer $n$ such that every red-blue colouring of the edges of $K_n$ contains a monochromatic $K_t$.

More generally, let $R(s,t)$ be the least $n$ forcing either a red $K_s$ or a blue $K_t$. We haveAssuming the two smaller numbers exist, colourand choose a vertex $v$. At least $R(s-1,t)$ of its incident edges are red, or at least $R(s,t-1)$ are blue. In the first case, the corresponding neighbourhood contains a red $K_{s-1}$, which extends with $v$ to a red $K_s$, or a blue $K_t$. The second case is symmetric. Thuswhich proves existence by induction.

Pascal's identity then gives the binomial upper bound for a Ramsey numberConsequently, for $t\ge 2$,