# 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?

• $d>n$ implies $d$ can't divide $n$. – Calvin Khor Sep 2 '15 at 2:31
• The question is formatted like that because it gives you a hint. – Jorge Fernández-Hidalgo Sep 2 '15 at 2:33
• Ah well. Fair enough :) – Calvin Khor Sep 2 '15 at 2:36
• It actually doesn't really help solve the problem to know that $d\not\mid n$ - it is both redundant and a red herring. – Thomas Andrews Sep 2 '15 at 3:46
• It does. The number of values so that $d<n$ is intimately related to the number of values of $d$ so that $d$ does not divide $n$ – Jorge Fernández-Hidalgo Sep 2 '15 at 5:03

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 $\frac{(2a_1+1)\cdots (2a_k+1)-1}{2}$