Given sequence $(x_n)$ satisfying $$ x_1 = 1, x_{n + 1} = x_n + \frac{1}{(\sqrt m – 1)x_n}, \forall n \ge 1$$
Find all positive integer $m$ $(m > 1)$ such that $$ \lfloor mx_{m^2}\rfloor = m^2$$
We have that $$x_1 = 1, x_2 = \dfrac{\sqrt m}{\sqrt m – 1}, x_3 = \dfrac{m + \sqrt m – 1}{\sqrt m(\sqrt m – 1)}, x_4 = \dfrac{m^2 + 3\sqrt{m^3} – 2m – 2\sqrt m + 1}{(\sqrt m – 1)\sqrt m(m – \sqrt m + 1)}, \cdots$$
There doesn't seem to be any patterns here.
It could also be observed that $m = 1$ is the obvious solution.
I have also been told that $m = 9$ is a solution.
I wonder which certain values of $m = p^2$ $(p \in \mathbb Z^+)$ satisfy the condition above.
Best Answer
Certainly not. For $m=4$, $\lfloor 4 x_4 \rfloor = 23$. It is true that $m=9$ works. Numerical experiments (using floating-point approximations as the exact values of $x_n$ rapidly get out of hand) seem to show that $x_{m^2}$ grows more slowly than $m$ as $m$ increases. I suspect that suitable bounds could prove this.
EDIT: Consider the iteration $x_{n+1} = x_n + \dfrac{a}{x_n}$ where $x_1 = 1$ and $0 < a \le 1$. Note that $$x_{n+1}^2 = x_n^2 + 2 a + \frac{a^2}{x_n^2} \le x_n^2 + 2 a + a^2$$
By induction, $x_n \le \sqrt{1 + (2a+a^2) (n-1)}$. In our case we have $a = \dfrac{1}{\sqrt m - 1}$, so $$ m x_{m^2} \le m\sqrt {1+ \left( \left( \sqrt {m}-1 \right) ^{-2}+2\, \left( \sqrt { m}-1 \right) ^{-1} \right) \left( {m}^{2}-1 \right) } $$ and this is less than $m^2$ if $m > (2 + \sqrt{2})^2 \approx 11.65$.
Conclusion: After eliminating other values $\le 11$, $m=1$ and $m=9$ are the only solutions.