
A 7. Let $A$ be a set of positive integers with the following properties:

(a) If $n$ is an element of $A$ then $n \leqslant 2018$.

(b) If $S$ is a subset of $A$ with $|S|=3$ then there are two elements $n, m$ of $S$ with $|n-m| \geqslant \sqrt{n}+\sqrt{m}$.

What is the maximum number of elements that $A$ can have?


Solution. Assuming $n>m$ we have

$$
\begin{aligned}
|n-m| \geqslant \sqrt{n}+\sqrt{m} & \Leftrightarrow(\sqrt{n}-\sqrt{m})(\sqrt{n}+\sqrt{m}) \geqslant \sqrt{n}+\sqrt{m} \\
& \Leftrightarrow \sqrt{n} \geqslant \sqrt{m}+1 .
\end{aligned}
$$

Let $A_{k}=\left\{k^{2}, k^{2}+1, \ldots,(k+1)^{2}-1\right\}$. Note that each $A_{k}$ can contain at most two elements of since if $n, m \in$ with $n>m$ then

$$
\sqrt{n}-\sqrt{m} \leqslant \sqrt{(k+1)^{2}-1}-\sqrt{k^{2}}<(k+1)-k=1
$$

In particular, since $\subseteq A_{1} \cup \cdots \cup A_{44}$, we have $|S| \leqslant 2 \cdot 44=88$.

On the other hand we claim that $A=\left\{m^{2}: 1 \leqslant m \leqslant 44\right\} \cup\left\{m^{2}+m: 1 \leqslant m \leqslant 44\right\}$ satisfies the properties and has $|A|=88$. We check property (b) as everything else is trivial.

So let $r, s, t$ be three elements of $A$ and assume $r<s<t$. There are two cases for $r$.

(i) If we have that $r=m^{2}$, then $t \geqslant(m+1)^{2}$ and so $\sqrt{t}-\sqrt{r} \geq 1$ verifying (b).

(ii) If we have that $r=m^{2}+m$, then $t \geqslant(m+1)^{2}+(m+1)$ and

$$
\begin{aligned}
\sqrt{t} \geqslant \sqrt{r}+1 & \Leftrightarrow \sqrt{(m+1)^{2}+(m+1)} \geqslant \sqrt{m^{2}+m}+1 \\
& \Leftrightarrow m^{2}+3 m+2 \geqslant m^{2}+m+1+2 \sqrt{m^{2}+m} \\
& \Leftrightarrow 2 m+1 \geqslant 2 \sqrt{m^{2}+m} \\
& \Leftrightarrow 4 m^{2}+4 m+1 \geqslant 4 m^{2}+4 m .
\end{aligned}
$$

So property (b) holds in this case as well.

## COMBINATORICS
