Maya Lists the Positive Divisors 
Maya lists all the positive divisors of $2010^2$. She then randomly selects two distinct divisors from this list. Let $p$ be the probability that exactly one of the selected divisors is a perfect square. The probability $p$ can be expressed in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

$2010^2 = 2^2 \cdot 3^2 \cdot 5^2 \cdot 67^2$
It has, $(3)(3)(3)(3) = 3^4 = 81$ total divisors. 
Now I need to find how many are perfect squares. It certainly has two, $1^2, 2010^2$. But this is complicated, any hints?

The perfect squares must have an even power of each prime factor.  Therefore, you could count the divisors of $2010$; the square of each of these is a square divisor of $2010^2$.
More precisely, since $2010^2$ has 4 distinct pairs of divisors, if $a$ is a square divisor of $2010^2$ and $2$ is a factor of $a$, then $2^2$ is also a factor of $a$.  Extending this idea, there are $2^4=16$ total square divisors of $2010^2$.
Edit (from comments):
Claim: Let $a^2\mid b^2$, then $a\mid b$.
Proof: We can prime factorize $a$ and $b$ as $a=p_1^{a_1}\cdots p_n^{a_n}$ and $b=p_1^{b_1}\cdots p_n^{b_n}$.  In this prime factorization, some of the powers $a_i$ may be zero.  Since $a^2\mid b^2$, it follows that $2a_i\leq 2b_i$.  Then, $a_i\leq b_i$, so it follows that $a\mid b$.
From this, it follows that the square divisors of $b^2$ are precisely the divisors of $b$, squared.
