Solving equations with floor function and square roots? 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?

Hint: $10^n-1$ is a multiple of $9$, so $\frac{2}{3}k$ is an integer, then:
$$
\begin{align}
\left\lfloor{\frac{2}{3}\sqrt{k(k+2)}}\right\rfloor = \frac{2}{3}k \;\;&\iff\;\; \frac{2}{3}k \le \frac{2}{3}\sqrt{k(k+2)} \lt \frac{2}{3}k \,+\, 1 \;\; \\
 &\iff\;\; \frac{4}{9}k^2 \le \frac{4}{9}(k^2+2k) \lt \frac{4}{9}k^2 + \frac{4}{3}k+1 \\
 &\iff\;\; 0 \le \frac{8}{9} k \lt \frac{4}{3}k+1 \\
\end{align}
$$
