Let $H_{n}=\{\sqrt{2 \pm \sqrt{2 \cdots \pm \sqrt{2}}}\}$, where the number of square roots is $n$. How many real elements does the set $\bigcup_{n=1}^{2000} H_{n}$ have?
I. solution. We will prove by complete induction on $n$ that if $1 \leq i \leq j \leq n$, then the elements of the sets $H_{i}$ and $H_{j}$ are positive real numbers less than 2 and there are no equal elements among them.

If $n=1$, then this is obvious. Now let $n>1$ and assume that the statement is true for sets with indices less than $n$. The set $H_{n}$ consists of numbers of the form $x_{n}=\sqrt{2 \pm x_{n-1}}$, where $x_{n-1} \in H_{n-1}$. By the induction hypothesis, $0 < x_{n-1} < 2$ and there are no equal elements in $H_{n-1}$. Then, as can be easily verified, $r_{n}=a_{1}\left(1+\frac{r_{n-1}}{2}\right)$, where $r_{n-1} \in R_{n-1}$.

Thus, $2 \sin r_{n} \cdot \frac{\pi}{4}=2 \sin a_{1}\left(1+\frac{r_{n-1}}{2}\right) \frac{\pi}{4}=2 a_{1} \sin \left(1+\frac{r_{n-1}}{2}\right) \frac{\pi}{4}$. It is known that $\sin \alpha= \pm \sqrt{\frac{1-\cos 2 \alpha}{2}}$, which can be written as $2 \sin \alpha=a_{1} \sqrt{2-2 \cos 2 \alpha}$ ($a_{1}= \pm 1$). Now $\alpha=\left(1+\frac{r_{n-1}}{2}\right) \cdot \frac{\pi}{4}$, so

$$
2 \sin r_{n} \cdot \frac{\pi}{4}=a_{1} \sqrt{2-2 \cos \left(\frac{r_{n-1}}{2}+1\right) \cdot \frac{\pi}{2}}
$$

Since $\cos \left(x+\frac{\pi}{2}\right)=-\sin x$,

$$
2 \sin r_{n} \frac{\pi}{4}=a_{1} \sqrt{2+2 \sin \frac{r_{n-1}}{2} \cdot \frac{\pi}{2}}=a_{1} \sqrt{2+2 \sin r_{n-1} \cdot \frac{\pi}{4}}
$$

Under the square root, by the induction hypothesis,

$$
2+a_{2} \sqrt{2+a_{3} \sqrt{2+\cdots+a_{n} \sqrt{2}}}
$$

stands, thus we have completed the proof of $(*)$.

Based on the solution by Péter Pallos (Fazekas M. Fóv. Gyak. Gimn., 10th grade)

Remark. If we introduce the polynomials $p_{1}(x)=x^{2}-2$ and $p_{n}(x)=p_{n-1}^{2}(x)-2=p_{1}\left(p_{n-1}(x)\right)$, it can be easily seen that the elements of the set $H_{n}$ and their negatives are the roots of the $2^{n}$-degree polynomial $p_{n}(x)$. These polynomials are closely related to the so-called Chebyshev polynomials. It is known that for a given natural number $n$, $\cos n x$ can be written as an $n$-degree polynomial in $\cos x$: $\cos 0 x=1=T_{0}(x)$, $\cos 1 x=\cos x$ so $T_{1}(x)=x$, $\cos 2 x=2 \cos ^{2} x-1$ so $T_{2}(x)=2 x^{2}-1$, and generally, for the $n$-degree $T_{n}$ Chebyshev polynomial,

$$
\cos n x=T_{n}(\cos x)
$$

The polynomial $p_{1}(x)$ is exactly $2 T_{2}\left(\frac{x}{2}\right)$, and it immediately follows that the $2^{n}$-degree polynomial $p_{n}(x)$ is nothing but $\frac{1}{2} T_{2^{n}}(2 x)$. This means that

$$
p_{n}(2 \cos x)=2 \cdot \cos 2^{n} x
$$

Based on this relationship, we can write $2^{n}$ numbers in trigonometric form, all of which are roots of the polynomial $p_{n}(x)$. Thus, we list all the roots of the polynomial, since the degree of $p_{n}(x)$ is exactly $2^{n}$. We have a justified hope for this because we know that the elements of the set $H_{n}$ and their negatives are the roots of the polynomial $p_{n}(x)$, and these numbers, as we have seen, fall between -2 and 2, and can thus be written in the form $2 \cos x$.

Thus, we seek those values of $x$ for which $2 \cos 2^{n} x=0$; if $x_{i}$ is such a number, then $2 \cos x_{i}$ is a root of $p_{n}(x)$.

$\cos 2^{n} x=0$ precisely when $2^{n} x=k \frac{\pi}{2}$, where $k$ is an odd integer, so the roots of $p_{n}(x)$ are the numbers of the form

$$
2 \cos k \cdot \frac{\pi}{2^{n+1}}
$$

where $k$ is odd. If $1 \leq k<2^{n+1}$, then $0<k \frac{\pi}{2^{n+1}}<\pi$, and since the cosine function is strictly decreasing on the interval $[0 ; \pi]$, there are no equal numbers among the $2^{n}$ numbers listed, thus we have obtained the roots of the polynomial $p_{n}(x)$. Half of these $2^{n}$ numbers are positive, those for which $k<2^{n}$, which immediately shows that the set $H_{n}$ has $2^{n-1}$ elements.

Finally, the fact that the sets $H_{n}$ and $H_{m}$ are disjoint directly follows from the above trigonometric form of these numbers. If $n \neq m$, $k$ and $l$ are odd, then the sum and the difference of the arguments $\frac{k \pi}{2^{n+1}}$ and $\frac{l \pi}{2^{m+1}}$ are of the form $\frac{p}{q} \pi$, where $q$ is a power of 2, and $p$ is odd. Since the cosines of two numbers are equal precisely when the sum or the difference of the numbers is an even multiple of $\pi$, equality cannot hold in this case.