Local Mathematical Competitions

Consider such a set $X\subseteq S(m,n)$. Clearly, $S(m,n)$ and thus a fortiori $X$ are finite sets. Then $$\sum _{\mathbf{x}\in X}{x}_k\ge \sum _{i=1}^{|X|}(i-1)=|X|(|X|-1)/2,\text{ for all }k=1,2,\dots ,m,$$ and so $$n|X|=\sum _{\mathbf{x}\in X}\sum _{k=1}^m{x}_k=\sum _{k=1}^m\sum _{\mathbf{x}\in X}{x}_k\ge m|X|(|X|-1)/2,$$ hence $n\ge m(|X|-1)/2$, therefore $|X|\le \lfloor \frac{2n}{m}\rfloor +1$. This bound is thus valid for all points, and we claim it is sharp in all cases,¹¹ so $N(m,n)=\lfloor \frac{2n}{m}\rfloor +1$.

Let us now, for $m=3$, build a model set $X$ with $|X|=\lfloor \frac{2n}{3}\rfloor +1$. When $n=3k-1$, the above expression evaluates at $2k$, and one can use the triplets $(0,k+1,2k-2)$, $(1,k+2,2k-4)$, ..., $(k-1,2k,0)$, $(k,0,2k-1)$, $(k+1,1,2k-3)$, ..., $(2k-1,k-1,1)$. When $n=3k$, the above expression evaluates at $2k+1$, and one can use the triplets $(0,k,2k)$, $(1,k+1,2k-2)$, ..., $(k,2k,0)$, $(k+1,0,2k-1)$, $(k+2,1,2k-3)$, ..., $(2k,k-1,1)$. When $n=3k+1$, the above expression evaluates at $2k+1$, and one can use the triplets $(0,k,2k+1)$, $(1,k+1,2k-1)$, ..., $(k,2k,1)$, $(k+1,0,2k)$, $(k+2,1,2k-2)$, ..., $(2k,k-1,2)$.

Denote now, in the general case, $n^{\prime }=\lfloor \frac{2n}{m}\rfloor$ and $m^{\prime }=m-2$. Then $N(2,n^{\prime })=n^{\prime }+1$. We will describe how to build a model inductively.

¹¹$N(2,n)=n+1=|S(2,n)|$, as trivially seen, while $N(2n,n)=2$, given by e.g. $(0,0,\dots ,0,1,1,\dots ,1)$ and $(1,1,\dots ,1,0,0,\dots ,0)$.

Now $\lfloor \frac{2(n-n^{\prime })}{m^{\prime }}\rfloor \ge \lfloor \frac{2n}{m}\rfloor$, since $\frac{2(n-n^{\prime })}{m^{\prime }}\ge \frac{2n}{m}$ is equivalent to $m(n-n^{\prime })\ge (m-2)n$, or $2n\ge m{n}^{\prime }$, or $n^{\prime }\le \frac{2n}{m}$, patently true given the definition of $n^{\prime }$, so the model for $N(m^{\prime },n-n^{\prime })$ yields enough elements that may be adjoined to those of the model realizing $N(2,n^{\prime })$, in order to create one for $N(m,n)$ (valid, since $m=m^{\prime }+2$ and $n=(n-n^{\prime })+n^{\prime }$).

For the quite interesting particular case $m=n$, a model set $X=\{\mathbf{x},\mathbf{y},\mathbf{z}\}$ with $|X|=\lfloor \frac{2n}{n}\rfloor +1=3$ is also easily built inductively. Start with $n=2$ and $\mathbf{x}=(0,2),\mathbf{y}=(2,0),\mathbf{z}=(1,1)$. Have $\mathbf{x}=(x_1,\dots ,x_k,2)$, $\mathbf{y}=(y_1,\dots ,y_k,0)$, $\mathbf{z}=(z_1,\dots ,z_k,1)$ already built, for some $1\le k<n-1$, and build $\mathbf{x}=(z_1,\dots ,z_k,0,2)$, $\mathbf{y}=(y_1,\dots ,y_k,1,0)$, $\mathbf{z}=(x_1,\dots ,x_k,2,1)$, until reaching the full $n$ coordinates.