Solve the following equation:

$$
\sqrt{x+1999 \sqrt{x+1999 \sqrt{x+1999 \sqrt{x+1999 \sqrt{2000 x}}}}}=x
$$
It is easily visible that 0 and 2000 are solutions to the equation. We will show that there are no other solutions.

Indeed, if $0\sqrt{x^{2}}=x$ then

$$
\sqrt{x+1999 \sqrt{2000 x}}>\sqrt{x+1999 x}=\sqrt{2000 x}>x
$$

and so on, the quantity under each square root on the left side of the equation is greater than $x$. Thus, the left side of the equation is greater than $x$, while the right side is equal to $x$, which is not possible.

If $x>2000$, then after $\sqrt{2000 x}<\sqrt{x^{2}}=x$, a similar reasoning shows that the left side of the equation is less than $x$, so we do not get a solution in this case either.

Remark. Taraza Busra (Budapest, ELTE Apáczai Cs. J. Gyak. Gimn., 12. o.t.) essentially followed the previous line of thought and generally proved that the

$$
\sqrt{\underbrace{x+n \sqrt{\cdots+n \sqrt{x+n \sqrt{(n+1) x}}}}_{k \text { terms }}}=x
$$

equation (where $n$ and $k$ are positive integers) has exactly two solutions, $x=0$ and $x=n+1$.