Is $\lfloor2\sqrt{n-\lfloor \sqrt{n} \rfloor}\rfloor <\lfloor2\sqrt{n} \rfloor$ true for $n\geq 0$?

Question

I have reasons to believe that $\lfloor2\sqrt{n-\lfloor \sqrt{n} \rfloor}\rfloor <\lfloor2\sqrt{n} \rfloor$ for $n\geq 0$. How could I go about proving (or disproving) this?

Answer 1

You could start by checking the inequality for small values of $n$. You would find that it is false already for $n=2$.

Answer 2

Hint. It is true if you replace $<$ with $\leq$ and use the fact that

Proposition. If $x\leq y \Rightarrow \lfloor x \rfloor \leq \lfloor y \rfloor$

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From that proposition, $\forall n\geq 0$ $$0\leq2\sqrt{n-\lfloor \sqrt{n} \rfloor} \leq 2\sqrt{n} \iff \lfloor \sqrt{n} \rfloor \geq 0$$ and as a result

$$\lfloor2\sqrt{n-\lfloor \sqrt{n} \rfloor}\rfloor \leq\lfloor2\sqrt{n} \rfloor $$

Answer 3

$$\lfloor2\sqrt x\rfloor$$ is a non-decreasing function so it is obvious that

$$\left\lfloor2\sqrt{n-\lfloor\sqrt n\rfloor}\right\rfloor\le \lfloor2\sqrt n\rfloor$$ holds.

The interesting cases are equality, i.e. $$\left\lfloor2\sqrt{n-\lfloor\sqrt n\rfloor}\right\rfloor=\lfloor2\sqrt n\rfloor.$$

The function $\lfloor2\sqrt x\rfloor$ remains the constant $m$ for

$$m^2\le4n\le(m+1)^2-1.$$

So we need

$$m^2\le 4(n-\lfloor\sqrt n\rfloor)\le m^2+2m.$$

As $$\left\lfloor\frac m2\right\rfloor\le\lfloor\sqrt n\rfloor\le\left\lfloor\frac {\sqrt{m(m+2)}}2\right\rfloor,$$ by addition and taking the tightest bounds,

$$m^2+4\left\lfloor\frac m2\right\rfloor\le 4n\le m^2+2m,$$ which is not void for even $m$.

Every $m=2k$, i.e. $n=k(k+1)$ makes the equality, as

$$k(k+1)-\left\lfloor\sqrt{k(k+1)}\right\rfloor=k^2$$ and

$$\left\lfloor2\sqrt{k(k+1)}\right\rfloor=\left\lfloor2\sqrt{k^2}\right\rfloor=2k.$$

There are no other solutions.