Suppose $x>30$ is a number satisfying $\lfloor x\rfloor \cdot \lfloor x^2\rfloor = \lfloor x^3\rfloor$. Prove that $\{x\}<\frac{1}{2700}$, where $\{x\}$ is the fractional part of $x$.

My heuristic is that $x$ needs to be "small": i.e. as close to $30$ as possible to get close to the upper bound on $\{x\}$, but I'm not sure how to make this a proof.

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    $\begingroup$ Set $a=\{x\}$, $b=\lfloor x \rfloor$. Then $x=a+b$, $x^2=a^2+2ab+b^2$, $x^3=\cdots$. Plug in and simplify both sides. $\endgroup$ – vadim123 Jul 14 '16 at 17:15
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    $\begingroup$ By the way, what you've called your "heuristic" is really your "intuition" or "conjecture". A heuristic is a way of approximately doing something, like "predicting the winner of an election via exit polls with a small sample is a good heuristic in all but the closest races." $\endgroup$ – John Hughes Jul 14 '16 at 17:19
  • $\begingroup$ vadim123 I'd got to that part, but I'm not sure how to proceed. $\endgroup$ – jlammy Jul 14 '16 at 17:33
  • $\begingroup$ $x = \lfloor x \rfloor + \{x\}$ find $x^2$ and $x^3$ in this format. Then what bounds must $\{x\}$ be in for the equation to balance. $\endgroup$ – Doug M Jul 14 '16 at 17:39

Let $\lfloor x \rfloor =y$ and $\{x\}=b$ Then $\lfloor x\rfloor \cdot \lfloor x^2\rfloor = \lfloor x^3\rfloor =y\lfloor y^2+2by+b^2 \rfloor= \lfloor y^3+3y^2b+3yb^2+b^3\rfloor$

One way this can happen is that $b$ is small enough that all the terms including $b$ are less than $1$, which makes both sides $y^3$. This requires $3y^2b \lt 1$, which gives $b \lt \frac 1{2700}$ as required. Now you have to argue that if $2by+b^2 \ge 1$ the right side will be too large.

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  • $\begingroup$ Could you flesh out the last bit? I'm getting $RHS\geq\lfloor(y+b)(y^2+1)\rfloor$ but don't know how to proceed? $\endgroup$ – jlammy Jul 14 '16 at 21:30
  • $\begingroup$ The only way to increase the left side is is $2by+b^2 \ge 1$. That puts a minimum on $b$ which depends on $y$. If it is just a little greater than $1$, the left side will be $y(y^2+1)=y^3+y$. Plug the minimum $b$ into the expression on the left and show it is greater than that. $\endgroup$ – Ross Millikan Jul 14 '16 at 22:23

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