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?

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 $n$.
