How many positive integers x and y satisfy the equation $x^y = (2016)^{2016}$? How many positive integers x and y satisfy the equation $x^y = (2016)^{2016}$ ? Explain your answer.
I started by factoring $2016$. I found the factors to be $36$, but I couldn't go further. 

We know that $2016 = 2^5\cdot 3^2\cdot 7^1$ and so $2016^{2016} = 2^{2016\cdot 5}\cdot 3^{2016\cdot 2}\cdot 7^{2016}$
We want $x^y$ to be equal to $2016^{2016}$.  It is clear then by the prime factorization of $2016$ and the fundamental theorem of arithmetic that $x$ must be of them form $x=2^{a}\cdot 3^b\cdot 7^c$, implying that $x^y=2^{ay}\cdot 3^{by}\cdot 7^{cy}$
Further, to have $2016^{2016}=x^y$ we will require that: $\begin{cases}ay=5\cdot 2016\\ by=2\cdot 2016\\ cy=2016\end{cases}$
Thus, we may simplify, $x^y = 2^{5cy}\cdot 3^{2cy}\cdot 7^{cy}$
Since we want $x$ and $y$ to be positive integers, this implies further that $c$ must be a positive integer.  We see that the number of solutions $(x,y)$ will be in bijection with the solutions to $cy=2016$ which are in bijection with the divisors of $2016$.
As there are $(5+1)\cdot (2+1)\cdot (1+1)=6\cdot 3\cdot 2 = 36$ divisors of $2016$, each of these divisors may play the role of $y$ and the corresponding value for $x$ will be $x=2^{2016\cdot 5/y}\cdot 3^{2016\cdot 2/y}\cdot 7^{2016/y}$ (which is clearly an integer since $y\mid 2016$)
