For a real number $x$, $\{x\}+\left\{x^{-1}\right\}=1$. Determine

$$
\left\{x^{2001}\right\}+\left\{x^{-2001}\right\}
$$

value.
Let's solve the problem for an arbitrary exponent $n$ instead of 2001.

Considering that the given and the desired expressions have a similar form, let's examine when $\{a\}+\left\{\frac{1}{a}\right\}$ can be an integer. Given $0 \leq\{a\}<1$ and $0 \leq\left\{\frac{1}{a}\right\}<1$, the sum can be either 1 or 0. The latter is only possible if both $a$ and $\frac{1}{a}$ are integers, i.e., $a = \pm 1$. Thus, $1 = \{x\} + \left\{\frac{1}{x}\right\} = x - [x] + \frac{1}{x} - \left[\frac{1}{x}\right]$, which implies that $x + \frac{1}{x}$ is an integer, but neither $x$ nor $\frac{1}{x}$ is an integer.

We will prove by induction on $n$ that for any $n$, $x^n + \frac{1}{x^n}$ is an integer. For $n=1$, the initial condition is satisfied, which is clearly true. Assume that for some $k$, the statement holds for all natural numbers up to $k$. Then, $\left(x + \frac{1}{x}\right) \cdot \left(x^k + \frac{1}{x^k}\right) = x^{k+1} + \frac{1}{x^{k+1}} + \frac{1}{x^{k-1}} + x^{k-1}$, and thus

$$
x^{k+1} + \frac{1}{x^{k+1}} = \left(x + \frac{1}{x}\right) \cdot \left(x^k + \frac{1}{x^k}\right) - \left(\frac{1}{x^{k-1}} + x^{k-1}\right)
$$

By the induction hypothesis, the right-hand side consists of the integers

$$
\left(x + \frac{1}{x}\right), \quad \left(x^{k-1} + \frac{1}{x^{k-1}}\right), \quad \left(x^k + \frac{1}{x^k}\right)
$$

Thus, $x^{k+1} + \frac{1}{x^{k+1}}$ is also an integer, which completes the proof. According to the proven statement (for all $n$),

$$
x^n + \frac{1}{x^n} - [x^n] - \left[\frac{1}{x^n}\right] = \left\{x^n\right\} + \left\{\frac{1}{x^n}\right\}
$$

is an integer.

We have seen that such a sum can only be an integer if it is 0 or 1; 0 only if $x^n = \pm 1$, i.e., $x = \pm 1$, but this possibility is excluded by the problem's conditions. Therefore, for every positive integer $n$, $\left\{x^n\right\} + \left\{x^{-n}\right\} = 1$.

We will show that there exists an $x$ that satisfies the initial condition. For example, let $x = \frac{3 + \sqrt{5}}{2}$, then

$$
\frac{1}{x} = \frac{2}{3 + \sqrt{5}} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}
$$

Their sum is an integer, specifically 3, but they themselves are not integers, so the sum of their fractional parts is 1.

Gábor Fehér (Budapest, Berzsenyi D. Gymnasium, 10th grade)

Roland Paulin (Fazekas M. Fôv. Gyak. Gymnasium and Secondary School, 8th grade) based on their solutions

Remarks. 1. The question arises as to which $x$ satisfy the given relationship in the problem. From the condition $x + \frac{1}{x} \in \mathbb{Z}$, it follows that $x + \frac{1}{x} = k$ (where $k \in \mathbb{Z}$). This leads to the quadratic equation $x^2 - kx + 1 = 0$. The solutions to this equation are numbers of the form $\frac{k \pm \sqrt{k^2 - 4}}{2}$, and the reciprocal of $\frac{k + \sqrt{k^2 - 4}}{2}$ is precisely $\frac{k - \sqrt{k^2 - 4}}{2}$. The obtained root provides a solution for every integer $|k| \geq 2$, but for $|k| = 2$, the root itself is an integer. This $x$ is not suitable, so $|k| \geq 3$ is necessary; moreover, it can be easily shown that $x$ cannot be an integer for other values of $k$.
2. Knowing the possible values of $x$, it is easy to see that for even $k$, $x^{2001} + \frac{1}{x^{2001}}$ is an integer, but for odd $k$, this is not so obvious. In this case, the same can be calculated using the induction seen in the solution (or by checking the coefficients of the 2002 terms in the sum). From this (as we saw in the solution), it follows that the sum of the fractional parts is also an integer, but not 0, so it is 1.