Solving equations with floor function and square roots?

Question

I was solving a problem to discover n and after I transformed the problem it gave me this equation:

\begin{equation*} \left\lfloor{\frac{2}{3}\sqrt{10^{2n}-1}}\right\rfloor = \frac{2}{3}(10^{n}-1) \end{equation*}

So I tried to simplify it by defining: \begin{equation*} k = 10^{n}-1 \end{equation*}

and was left with: \begin{equation*} \left\lfloor{\frac{2}{3}\sqrt{k(k+2)}}\right\rfloor = \frac{2}{3}k \end{equation*}

But I can't get past that. Can anyone help me?

Answer 1

Hint: $10^n-1$ is a multiple of $9$, so $\frac{2}{3}k$ is an integer, then:

$$ \begin{align} \left\lfloor{\frac{2}{3}\sqrt{k(k+2)}}\right\rfloor = \frac{2}{3}k \;\;&\iff\;\; \frac{2}{3}k \le \frac{2}{3}\sqrt{k(k+2)} \lt \frac{2}{3}k \,+\, 1 \;\; \\ &\iff\;\; \frac{4}{9}k^2 \le \frac{4}{9}(k^2+2k) \lt \frac{4}{9}k^2 + \frac{4}{3}k+1 \\ &\iff\;\; 0 \le \frac{8}{9} k \lt \frac{4}{3}k+1 \\ \end{align} $$

Answer 2

Note that $$-8(10^n-1)\ \le\ 0\ <\ 4\cdot10^n+5$$ is true for all $n\in\mathbb Z^+$. In other words $$-8\cdot10^n+4\ \le\ -4\ <\ 4\cdot10^n+1$$ $$\iff\ 4(10^n-1)^2\ \le\ 4(10^{2n}-1)\ <\ (2\cdot10^n+1)^2$$ $$\iff\ 2(10^n-1)\ \le\ 2\sqrt{10^{2n}-1}\ <\ 2\cdot10^n+1$$ $$\iff\ \frac23(10^n-1)\ \le\ \frac23\sqrt{10^{2n}-1}\ <\ \frac{2\cdot10^n+1}3=\frac23(10^n-1)+1$$ $$\iff\ \left\lfloor\frac23\sqrt{10^{2n}-1}\right\rfloor\ =\ \frac23(10^n-1)$$ since $10^n-1$ is a multiple of 9 and so is divisible by 3. Hence your equation is true for all non-negative integers $n$.