My question is as is in the title. My way of trying to solve it was to form the inequality $ 0 \geq (k-2)r + 1 -k $ by saying that $ r^2 > (r-1)^2 + k(r-1) $ for some $k > 0$. The obvious solutions are perfect squares but for integers that aren’t perfect squares I believe there is 2 solutions for n in my inequality $ r^2 > n >(r-1)^2$ but I don’t know how to prove it with my inequality.
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wrote a little program, the $n$ that work seem to be $k^2 - 1, ; k^2, ; k^2 + k $ where the last one is the geometric mean of consecutive squares, same as double the triangular numbers. – Will Jagy Dec 12 '23 at 23:41
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@WillJagy FYI, your $k^2-1$ is the same as $(k-1)^2 + 2(k-1)$, so in general the values which work are $k^2$, $k^2+k$ and $k^2+2k$, as indicated in this answer to a basically duplicate question. Note the method used there is basically how I also approached solving this myself. – John Omielan Dec 12 '23 at 23:53
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1Does this answer your question? Find all positive integers such that $\lfloor{\sqrt{n}\rfloor} \mid n$ - found through using an Approach0 search. As shown in the answer there, with $k=\lfloor \sqrt{n}\rfloor$, the solutions for $n$ are $k^2$, $k^2+k$ and $k^2+2k$, i.e., $2$ solutions for $n$ between consecutive perfect squares, as you surmised. Also, FYI, the ... – John Omielan Dec 13 '23 at 00:02
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(cont.) Approach0 search also found quite a few other similar questions, e.g., How many numbers such $n$ are there that $n<100,\lfloor{\sqrt{n}} \rfloor \mid n$, How many positive integers $N \le 10^6$ such that $\lfloor \sqrt N \rfloor|N$?, Methodical way of finding the number of $n$ such that $\lfloor\sqrt n\rfloor\mid n$ and $\lfloor \sqrt{n+1}\rfloor\mid n+1$, Simple number theory in one unknown, Find the number of $m$ for which holds that $\lfloor{\sqrt m}\rfloor$ divides $m$, etc. Finally, welcome to Math SE. – John Omielan Dec 13 '23 at 00:08