Find all sets $X$ consisting of at least two positive integers such that for every pair $m, n \in X$, where $n>m$, there exists $k \in X$ such that $n=m k^{2}$.

Answer: The sets $\left\{m, m^{3}\right\}$, where $m>1$.
Let $X$ be a set satisfying the condition of the problem and let $n>m$ be the two smallest elements in the set $X$. There has to exist a $k \in X$ so that $n=m k^{2}$, but as $m \leq k \leq n$, either $k=n$ or $k=m$. The first case gives $m=n=1$, a contradiction; the second case implies $n=m^{3}$ with $m>1$.

Suppose there exists a third smallest element $q \in X$. Then there also exists $k_{0} \in X$, such that $q=m k_{0}^{2}$. We have $q>k_{0} \geq m$, but $k_{0}=m$ would imply $q=n$, thus $k_{0}=n=m^{3}$ and $q=m^{7}$. Now for $q$ and $n$ there has to exist $k_{1} \in X$ such that $q=n k_{1}^{2}$, which gives $k_{1}=m^{2}$. Since $m^{2} \notin X$, we have a contradiction.

Thus we see that the only possible sets are those of the form $\left\{m, m^{3}\right\}$ with $m>1$, and these are easily seen to satisfy the conditions of the problem.