1. ## difficult problem

Let "n= 2^{31}3^{19}. How many positive divisors of n^2 are less than n but
do not divide n?

2. "Raise" is an odd notation! Powers are commonly written as "^". I think you mean "n= 2^{31}3^{19}" or, better, using LaTex, " $n= 2^{31}3^{19}$". $n^2$ then is $n^2= 2^{62}3^{38}$. To be a divisor of that but not of n itself, a number must be divisible by either $2^{32}$ or $3^{20}$.