I have been asked to prove that if an integer $n > 1$ has no integer divisors satisfying $1<d<\sqrt{n}$, then $n$ is prime. I had issue with the question, because $n = 9$ is a counterexample to this, and so, really, I think I should be asked to prove that "if an integer $n > 1$ has no integer divisors satisfying $1 < d \leq \sqrt{n}$, then $n$ is prime." But, we've asked, and our professor insists that it's supposed to be strict inequality...
I know how to prove the case for $1 < d \leq \sqrt{n}$:
Suppose $n>1$ is not prime. Then, $\exists $ integers $a,d$ such that $n = a\cdot d$. Suppose, WLOG, that $d<a$. Then, $\displaystyle d = > \frac{n}{a} \leq \frac{n}{d} \, \implies \, d \leq \frac{n}{d} \, > \implies \, d^{2} \leq n \, \to \, d \leq \sqrt{n}$.
But, as I said, our professor insists we're proving this for $1 < d <\sqrt{n}$...
Am I wrong? Is $9$ not a counterexample for this, and if so, is there a way to prove this for $1<d<\sqrt{n}$?
Your professor is indeed wrong, and there are more counter-examples:
Consider $n = p^2$ for some prime $p$. The only divisors of $n$ are $1, p$ and $p^2$, and none of these lie strictly between $1$ and $\sqrt{n} = p$. Yet $n$ is not prime.