Prove that for any $n$ having at least two distinct prime factors, there exists a perfect square between $n$ and $\sigma(n)$ I came up with a new problem, described as follows: Prove that for any $n$ having at least two distinct prime factors, there exists at least one perfect square between $n$ and $\sigma(n)$
I will post my own solution for this problem, (if any part of it was wrong please notify me) but it is based on the inequality I have proved in here.
I would appreciate any help with the following:


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*Is there any other way to prove the statement, for example without using that inequality?

*Can we generalize the statement, so that we can say with $n$ having at least $t$ prime factors there is at least $s$ perfect squares between $n$ and $\sigma(n)$ for all $n>m$ for some natural number $m$?

 A: As $n$ has at least two distinct prime factors, say $p$ and $q$, we can say the number of $n$'s divisors is more than $4$ which are $(1,p,q,pq)$, so $d(n)\geq4$. Based on this inequality, we can say $\sigma(n)\geq\sqrt n(d(n)-2)+n+1\geq 2\sqrt n+n+1=(\sqrt n+1)^2$$\Rightarrow$$\sqrt{\sigma(n)}\geq\sqrt n+1$$\Rightarrow$$\sqrt{\sigma(n)}-\sqrt n\geq1$. One can easily prove that this inequality holds only when there is a perfect square between $n$ and $\sigma(n)$ (I) OR both $n$ and $\sigma(n)$ are consecutive perfect squares (II).Case (I) is clearly our problem, so we should only disprove case (II). Assume case (II) is true. Since $n$ and $\sigma(n)$ are consecutive perfect squares, we can say $\sigma(n)=n+2\sqrt n+1$ which is the equality case of the first-mentioned inequality. As mentioned in the linked question, the equality hold either when $n$ is a prime or it is the square of a prime, and both of them don't satisfy our assumption that $n$ has two distinct prime factors. So the problem is proven completely.
