Let $n=2^{31}3^{19}.$  How many positive [integer](https://artofproblemsolving.com/wiki/index.php/Integer) [divisors](https://artofproblemsolving.com/wiki/index.php/Divisor) of $n^2$ are less than $n_{}$ but do not divide $n_{}$?
We know that $n^2 = 2^{62}3^{38}$ must have $(62+1)\times (38+1)$ [factors](https://artofproblemsolving.com/wiki/index.php/Factor) by its [prime factorization](https://artofproblemsolving.com/wiki/index.php/Prime_factorization). If we group all of these factors (excluding $n$) into pairs that multiply to $n^2$, then one factor per pair is less than $n$, and so there are $\frac{63\times 39-1}{2} = 1228$ factors of $n^2$ that are less than $n$. There are $32\times20-1 = 639$ factors of $n$, which clearly are less than $n$, but are still factors of $n$. Therefore, using complementary counting, there are $1228-639=\boxed{589}$ factors of $n^2$ that do not divide $n$.