# 2006 Romanian NMO Problems/Grade 7/Problem 4

## Problem

Let be a set of positive integers with at least 2 elements. It is given that for any numbers , we have , where by we have denoted the least common multiple of and . Prove that the set has *exactly* two elements.

*Marius Gherghu, Slatina*

## Solution

We first show that is finite; for the sake of contradiction, suppose that is infinite. Let be the smallest element of . Let such that and let . Then is an integer for arbitrarily large ; but if . Therefore is finite.

Let with ; let and and . Then . Suppose and let be a prime dividing ; divides if and only if divides . But divides neither because and are relatively prime. Thus and and for all such that .

Let be the smallest element and let be the largest element of . Since , we have and divides and so either or and equals or . Suppose there exists some such that . If , then , contradiction. If , we can without loss of generality assume that is the second largest element of . Then so and which is a contradiction since .