Find the smallest integer that is divisible by exactly $X$ perfect squares. Is there a method to find the smallest integer divisible by exactly $X$ perfect squares?
Example: find the smallest positive integer divisible by exactly 2015 perfect squares.
I've been trying to figure this out, but I haven't made much progress... Please help!

First of all, there won't be a simple formula.
If $X$ is prime, the smallest number is $2^{2X-2}$. So the smallest number divisible by example five perfect squares is $2^{8}=256$. 
More generally, it will depend on factorizations of $X$.
In general, if $N$ is the smallest integer with exactly $X$ positive integer divisors, then $N^2$ will be the smallest integer with exactly $X$ perfect square divisors.
The number of divisors of $N$ is in terms of the prime factorization of $N$. So if $$N=p_1^{a_1}p_2^{a_2}\cdots p_n^{a_n}$$ then the number of divisors is:
$$\tau(N) = (a_1+1)(a_2+1)\cdots(a_n+1)$$
So lets try the example $X=15$. Then either $N=p_1^{14}$ or $N=p_1^{4}p_2^{2}$. In this case $2^43^2=144$ is the smallest number with exactly $15$ divisors, o $144^2$ is the smallest with number with exactly $15$ perfect square divisors.
It gets harder with even more facts. Try $X=12$. Then there are many different ways to factor 
$$\begin{align}X&=24\\&=12\cdot 2
\\&=8\cdot 3
\\&=6\cdot 4
\\&=6\cdot 2\cdot 2
\\&=4\cdot 3\cdot 2 
\\&= 3\cdot 2\cdot 2\cdot 2\end{align}$$
Which yields prospects:
$$N=2^{23},2^{11}3^1, 2^{7}3^2, 2^{5}3^{3}, 2^{5}3^15^1, 2^33^25^1, 2^23^15^17^1$$
Here, the best value is $2^33^25^1=360$, and $N^2=360^2$ is your answer.
