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
.