Can two integers have the same product of divisors? If $x$ is a positive integer and $y$ is a positive integer, can the product of the divisors of $x$ equal the product of the divisors of $y$ for some arbitrary $x$ and $y$? (The product of the divisors includes itself; the product of divisors of $4$ would be $1\cdot2\cdot4=8$)
 A: No, it is not possible for the product of the divisors of $x$ to equal the product of the divisors of $y$, except of course when $x=y$.
First, a derivation of the well-known formula for the product of the divisors of an integer.  Let $\tau(n)$ be the number of divisors of $n$.  Write all the divisors of $n$ in a row in increasing order, and, below each divisor $d$, write $n/d$:
$$
\begin{array}{ccccc}
d_1&d_2&d_3&\ldots&d_{\tau(n)}\\
n/d_1&n/d_2&n/d_3&\ldots&n/d_{\tau(n)}
\end{array}
$$
The second row also consists of the $\tau(n)$ divisors of $n$, this time in decreasing order, so the product of each row is $\prod_{d\vert n} d$.  The product of each column is $n$.  Hence the product of all the entries in both rows is $(\prod_{d\vert n} d)^2 = n^{\tau(n)}$.  This gives the desired formula:
$$
\prod_{d\vert n} d = n^{\tau(n)/2}.
$$
The question is then when it is possible that $x^{\tau(x)/2} = y^{\tau(y)/2}$.  It is clear from this formula that any prime divisor $p$ of $x$ must also divide $y$, and vice versa.  So the prime factorizations of $x$ and $y$ may be written as
$$
\begin{align}
x &= p_1^{e_1}\, p_2^{e_2}\, \cdots\, p_k^{e_k}\\
y &= p_1^{f_1}\, p_2^{f_2}\, \cdots\, p_k^{f_k}
\end{align}
$$
Looking at the factors of any $p_i$ in $x^{\tau(x)/2} = y^{\tau(y)/2}$, one sees that $e_i\cdot\tau(x)/2 = f_i\cdot\tau(y)/2$, or $e_i/f_i = \tau(y)/\tau(x)$.  If this ratio is $1$, then all the corresponding exponents are equal and $x = y$.
If the ratio is not $1$, then one of $x$ and $y$ has larger exponents throughout its prime factorization.  That number then has more divisors, so the product of those divisors is larger.
