2006 Romanian NMO Problems/Grade 7/Problem 4

Problem

Let $A$ be a set of positive integers with at least 2 elements. It is given that for any numbers $a>b$ , $a,b \in A$ we have $\frac{ [a,b] }{ a- b } \in A$ , where by $[a,b]$ we have denoted the least common multiple of $a$ and $b$ . Prove that the set $A$ has exactly two elements.

Marius Gherghu, Slatina

Solution

Suppose that $A$ is infinite. Let $a \in A$ be the smallest element of $A$ . Let $n \in A$ such that $n > a$ and let $g = GCD(n,a)$ . Then $\frac{[n,a]}{n-a} = \frac{na}{g(n-a)} = \frac{a}{g} + \frac{a^{2}}{g(n-a)}$ is an integer for arbitrarily large $n$ ; but $\frac{a^{2}}{g(n-a)} < \frac{a^{2}}{n-a} < 1$ if $n > a^{2}+a$ . Therefore $A$ is finite.

Let $x,y \in A$ with $x > y$ ; let $g = GCD(x,y)$ and $x_{0} = x/g$ and $y_{0} = y/g$ . Then $\frac{[x,y]}{x-y} = \frac{xy}{g(x-y)} = \frac{x_{0}y_{0}}{x_{0}-y_{0}}$ . Suppose $x_{0} - y_{0} > 1$ and let $p$ be a prime dividing $x_{0} - y_{0}$ ; $p$ divides $x_{0}$ if and only if $p$ divides $y_{0}$ . But $p$ divides neither because $x_{0}$ and $y_{0}$ are relatively prime. Thus $x_{0} - y_{0} = 1$ and $g = x-y$ and $\frac{xy}{(x-y)^{2}} \in A$ for all $x,y \in A$ such that $x > y$ .

Let $b$ be the smallest element and let $a$ be the largest element of $A$ . Since $\frac{ab}{(a-b)^{2}} \in A$ , we have $b \le \frac{ab}{(a-b)^{2}} \le a \implies b \le (a-b)^{2} \le a$ so $(a-b)^{2}$ equals $a$ or $b$ . Suppose there exists some $n \in A$ such that $b < n < a$ . If $(a-b)^{2} = b$ , then $\frac{[n,a]}{a-n} = \frac{an}{(a-n)^{2}} > \frac{ab}{(a-b)^{2}} = a$ , contradiction. If $(a-b)^{2} = a$ , then $\frac{[n,b]}{n-b} = \frac{bn}{(n-b)^{2}} < \frac{ab}{(a-b)^{2}} = b$ , contradiction.

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2006 Romanian NMO Problems/Grade 7/Problem 4

Problem

Solution

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