## Problem 1.

Consider the set $A$ of four-digit numbers that are at most equal to 2014. Determine the maximum number of elements of a subset of $A$ that contains only perfect squares, any two of which are coprime.
Solution. If $n^{2} \in A$, then $n \in\{32,33,34, \ldots, 44\}$ $3 p$

To meet the conditions stated, among the perfect squares in $A$, we will retain one perfect square that is a multiple of 4, one perfect square that is a multiple of 9 but not of 4, one perfect square that is a multiple of 25 but not of 4 or 9, etc. $3 p$

An example can be: $32^{2}, 33^{2}, 35^{2}, 37^{2}, 41^{2}$ and $43^{2}$.

The maximum number of elements is 6

$1 p$