Partitioning the set of divisors into two sets with equal product Inspired with this PROBLEM I am interested in those natural numbers that the set of their divisors can be partitioned into two sets with equal product. For example we can decompose divisors of $8$ into $\lbrace 1,8\rbrace$ and $\lbrace 2,4\rbrace$. Is the sequence of this numbers well-known? Is there any characterization for them ? Any suggestion would be helpful.  
 A: Gerry Myerson comments that $n$ can only satisfy this condition if it is a 4th power, or has at least two prime factors with odd multiplicities, or has at least one prime factor with multiplicity 3 modulo 4. This is also sufficient.
Case 1: If $n$ has two prime factors with odd multiplicities, say $p^a$ and $q^b$, then one of the sets in your partition is the set of divisors which have (a power of $p$ which ranges from 0 through $\frac{a-1}{2}$) XOR (a power of $q$ which ranges from 0 through $\frac{b-1}{2}$).
Case 2: If $n$ has at least one prime factor with multiplicity 3 modulo 4, say $p^a$, then one of the sets in your partition is the set of divisors which have a power of $p$ which is 0 or 3 (mod 4). Note that in this case, both of the sets in your partition contain the same number of divisors.
Case 3: If $n$ is a perfect fourth power, let its prime factors be $p_1<p_2<...<p_k$. Then call one of the sets in your partition $S$. A divisor is in $S$ if and only if it is $1$, or its lowest prime factor has multiplicity 0 or 1 (mod 4).
You can check that all of these constructions work out. It's easiest to think about this problem by trying to find a "balanced" partition of the lattice of $n$'s divisors.
