Number of divisors $d$ of $n^2$ so that $d\nmid n$ and $d>n$ I just wanted to share this nutshell with you guys, it is a little harder in this particular case of the problem:

Find the number of divisors $d$ of $a^2=(2^{31}3^{17})^2$ so that $d>a$. What is the solution for the general case?


I think the number of divisors is (17+1)(31+1)-1=575.
d/a^2 and d>a.
{1,2,2^2,...,2^31}{1,3,3^2,...,3^17}={(1,1),(1,3),(1,3^2),...,(1,2),(1,2^2),...}
From (e,f) then ef is one divisor of a.
d=ef*a is devisor of a^2.
