## December 2009 solution

#### Part 1:

I'm using here a proof sent to me by James Ge.

Let us assume on the contrary.

First, note that both sets need to have infinite size. If one has a finite size, choose the smallest two elements from the other set that are larger than the maximum of the first set, as well as their power. These form a triplet that violates the conditions of construction.

So, we can choose any number of elements from each set, and we can choose elements to be arbitrarily large. Specifically, let us choose x from set A and y from set B, and let us assume w.l.o.g. that xy is in A. Also, let us pick an element r from A that is larger than xy.

In order not to create a triplet entirely in the same set, it can be deduced that rx and rxy are both in B. However, this leads to the triplet a=rx, b=y and c=rxy, all in set B - a contradiction.

It is left to the reader to verify that all triplets constructed in this proof are composed of three distinct elements.

#### Part 2:

Perhaps the most straightforward construction to solve this problem is to admit primes into one set and composites into the other. Clearly, a prime can never be represented as a non-trivial integer power.