Condition (b) rewrites to $|\sqrt{m}-\sqrt{n}| \ge 1$. Now, partition the set $T = \{ 1,2, \dots, 2018\}$ according to the floor of the square of its elements, i.e. $T_1 = \{ 1,2,3 \}, T_2 = \{ 4,5,6,7,8 \}, \dots, T_{44} = \{ 1936, 1937, \dots, 2018\}$. If there are three elements $a_1,a_2,a_3$ of $A$ who are in the same partition, then the maximal value of $|\sqrt{a_i}-\sqrt{a_j}|$ would be less than 1, violating (b). Therefore, there are at most two elements of $A$ in each partition. So, $|A| \le 2 \times 44 = 88$. For the construction, pick $A = \{ t^2, t^2+t | 1\le t \le 44, t \in \mathbb{Z} \}$. Take arbitrary $a_1<a_2<a_3\in A$. To show it works: If all of them are in different partition, then $|\sqrt{a_3}-\sqrt{a_1}| > 1$. If two of them are in one partition, bash cases and the observations $|\sqrt{(t+1)^2}-\sqrt{t^2}| \ge 1$ and $|\sqrt{(t+1)^2+t+1}-\sqrt{t^2+t}| \ge 1$ will do the work.

This problem was proposed by Demetres Christofides (Cyprus).

How did you rewrite it assqrt(m)-sqrt(n)>=1.

fabijan_cikac_123 wrote: How did you rewrite it assqrt(m)-sqrt(n)>=1. If we divide both sides of $$|n-m|\ge\sqrt{n}+\sqrt{m}$$by $\sqrt{n}+\sqrt{m}$, we get $$|\sqrt{n}-\sqrt{m}|\ge 1.$$

Steve12345 wrote: Let $A$ be a set of positive integers satisfying the following : $a.)$ If $n \in A$ , then $n \le 2018$. $b.)$ If $S \subset A$ such that $|S|=3$, then there exists $m,n \in S$ such that $|n-m| \ge \sqrt{n}+\sqrt{m}$ What is the maximum cardinality of $A$ ? Although it is A7, I think it is the easiet of Algebra. (1 of 2 which I could solve ) Step by step motivation 1 Write everything in your own language. I am thinking $a$ like "Every element of this set is equal to or smaller than $2018$" and $b$ like "Every subset of $A$ which has three elements contains two $m, n$ such that $|n-m| \ge \sqrt{n}+\sqrt{m}$" 2 I believe it is not hard to see $|\sqrt{m}-\sqrt{n}| \ge 1$. (Just the difference of two squares) 3 In my opinion the most important part is to figure out how to arrange $T_i$s. This is a well-known thing and if you struggle about squareroots and their differences, it is important to divide the set $K=\{1, \, ,2 ... \, 2018 \}$