Prove that values of $x$ satisfying $\lfloor{x^2\rfloor}=\left(\lfloor{x\rfloor}\right)^2$ is $(0,\sqrt{2})\cup \mathbb{Z}$

My try:

Its trivial that every integer satisfies the given equation.

Now if $x$ is not an integer we have $x=m+f$ where $m \in \mathbb{Z}$ and $0 \lt f \lt 1$

We have $\lfloor{x\rfloor}=m$

We have $$\lfloor{(m+f)^2\rfloor}=m^2$$


$$\lfloor{f^2+2mf\rfloor}=0$$ $\implies$

$$0 \le f^2+2mf \lt 1 \tag{1}$$

As per the comments, i understood my mistake.

Clearly $m$ cannot be negative integer since:

$$f^2+2mf \lt 0$$ which contradicts$(1)$

So $m$ is a non negative integer and for all such $m$, we have $f^2+2mf \ge 0$

Also from $(1)$ we have $$f^2+2mf \lt 1$$ $\implies$

$$f \in (0, \sqrt{m^2+1}-m)$$

Hence the final solution set is:

$$x \in [m, \sqrt{m^2+1})$$ $\forall$ non negative integers $m$

  • 1
    $\begingroup$ Is that really true? What about x=3.1? Or am I overlooking something? $\endgroup$ – Martin R Mar 21 at 18:09
  • $\begingroup$ Clearly $m=3,f=.1$ satisfy the last condition. Another easy scenario is $f=0$ $\endgroup$ – lab bhattacharjee Mar 21 at 18:12
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    $\begingroup$ For $x=3.1$ you have $\lfloor{x^2\rfloor}=\lfloor{9.61\rfloor}= 9 = \left(\lfloor{x\rfloor}\right)^2$ $\endgroup$ – Martin R Mar 21 at 18:15

Clearly $f=0$ is an obvious solution

Otherwise $0<f<1$

$$f^2+2mf<1\implies f<\sqrt{1+m^2}-|m|$$

$$f(f+2m)\ge0\iff f+2m\ge0\iff2m\ge-f\ge -1\iff m\ge0$$ as $m$ is an integer


$\newcommand{\fl}[1]{\left \lfloor #1\right \rfloor}$

Your statement is not true.

To see this, note that the function $g(x) = \fl{x^2} - [\fl{x}]^2$ takes only integer values, but is also zero at every integer.

However, also note that the function $\fl{\cdot}$ is right continuous, because for any point $l$, we can find a right neighbourhood of $l$ having the same floor as $l$, by keeping the neighbourhood small enough so that we don't include the integer after $l$. Then within this neighbourhood, the floor function will be constant, thus satisfying right continuity at $l$.

With compositions and so on preserving right continuity, one sees that $g$ is right continuous.

Now, by right continuity at each integer, with say $\epsilon = \frac 12$, you can find a right neighbourhood of each integer such that the value of $g(x)$ is within half of zero. However, $g(x)$ takes only integer values so this forces $g(x) = 0$ in that interval!

In short, to the right of every integer we can find an interval in which the above equality holds. Trivially this shows that the set of zeros given does not suffice.


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