Note that , $M(n)=\frac{\tau(n)-1}{2}$. So we just have to find the number satisfying , $1 \le n \le 2013$ having maximum roots in range $1 \le n \le 2013$ And its trivial now

What is the $\tau(n)$?

shivangjindal wrote: Note that , $M(n)=\frac{\tau(n)-1}{2}$. So we just have to find the number satisfying , $1 \le n \le 2013$ having maximum roots in range $1 \le n \le 2013$ And its trivial now I think $M(n)=\tau(n)-1$ and I found that $\max M(n)=35(1 \le n \le 2013)$ but I can't find all $n$. The proof of $\max M(n)=35$ is like this: First of all, it's easy to prove $M(n)=\tau(n)-1$. Assume $a_0$ is the smallest positive integer such that $M(a_0)$ maximizes. Let $a_0=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ and let $(q_n)$ be the increasing sequence of all prime numbers. So $p_i \ge q_i$. If $p_i>q_i$, we consider $b_0=p_1^{a_1}\cdots q_i^{a_i}\cdots p_k^{a_k}$; then $b_0<a_0$ and $M(b_0)=M(a_0)$, contradiction. Hence $p_i=q_i \forall i$. However, $q_1q_2q_3q_4q_5=2310>2013$ so $k \le 4$. The left work is try $k=1,2,3,4$.

If I'm not mistaken, it should be $M(n)=\frac{\tau(n^2)-1}{2}$. This would give $121$ as the maximum with $n=1680$. Let $c=\frac{ab}{a+b}$ or $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$. Clearly, $c<a$. Doing the usual algebra, we get $(a+b)(a-c)=a^2$. So $a-c|a^2$, and there are $\tau(a^2)$ divisors of $a^2$ and since $a-c=a$ is impossible, we have $\tau(a^2)-1$ divisors left. It is clear that half of these are less than $a$ and half are more than $a$, thus obtaining the formula shown. From there all it takes is a little organized trial and error to get the answer.

guys i can't understand here there are 2 solutions i don't which one is correct who can explain me correct solution please

The answer is M(n)=39 with n=1680

The answer is 112 for n=1680