a) We claim that for any numbers $p_1$ , $p_2$ , ... $p_n$ , $p_1^{n!}, p_2^{n!}, ... p_n^{n!}$ will satisfy the condition, which holds for any number $n$ .
Since $\sqrt[n] ab = \sqrt[n] a * \sqrt[n] b$ , we can separate each geometric mean into the product of parts, where each part is the $k$ th root of each member of the subset and the subset has $k$ members.
Assume our subset has $k$ members. Then, we know that the $k$ th root of each of these members is an integer (namely $p^{n!/k}$ ), because $k \leq n$ and thus $k | n!$ . Since each root is an integer, the geometric mean will also be an integer.
b) If we define $q$ as an arbitrarily large number, and $x$ and $y$ as numbers in set $S$ , we know that ${\sqrt[q]{\frac{x}{y}}}$ is irrational for large enough $q$ , meaning that it cannot be expressed as the fraction of two integers. However, both the geometric mean of the set of $x$ and $q-1$ other arbitrary numbers in $S$ and the set of $y$ and the same other $q-1$ numbers are integers, so since the other numbers cancel out, the geometric means divided, or ${\sqrt[q]{\frac{x}{y}}}$ , must be rational. This is a contradiction, so no such infinite $S$ is possible.
-aops111 (first solution dont bully me)