Number of divisors $d$ of $n^2$ so that $d\nmid n$ and $d>n$

Question

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?

Answer 1

General question: Number of divisors of $n^2$ that are greater than $n$

Let $n=p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}$ (and hence, $n^2=p_1^{2a_1}p_2^{2a_2}\dots p_k^{2a_k}$) where $p_i$'s are primes and $a_i$'s are positive integers. The number of divisors of $n$ is $(a_1+1)\cdots (a_k+1)$. If $n^2=a\cdot b$, then wlog $a\le n \le b$ (every divisor of $n^2$ smaller than $n$ has a counterpart which is greater than $n$). Omitting the divisor $n$, the number of divisors of $n^2$ which are greater than $n$ are $\frac{(2a_1+1)\cdots (2a_k+1)-1}{2}$

Answer 2

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.