Solve the equation $\{3 x\}^{2}+\{x\}^{2}=1$.

Note: The notation $\{x\}$ represents the fractional part of $x$.
Solution. Since the values of the functions $\{3 x\}$ and $\{x\}$ are the same at $x$ and $x+1$, it is sufficient to examine the equation in the interval $0 \leq x<1$. We distinguish three cases.

1) If $0 \leq x<\frac{1}{3}$, then the function is $(3 x)^{2}+x^{2}$, that is, the equation

$$
9 x^{2}+x^{2}=1
$$

needs to be solved. In this case, the solution is $x=\frac{1}{\sqrt{10}}$, and this indeed falls within the given interval.

2) If $\frac{1}{3} \leq x<\frac{2}{3}$, then the function is $(3 x-1)^{2}+x^{2}$, and the equation is

$$
9 x^{2}-6 x+1+x^{2}=1
$$

from which $10 x^{2}=6 x$ and $x=\frac{3}{5}$. It is easy to verify that this also falls within the specified interval.

3) If $\frac{2}{3} \leq x<1$, then the function is $(3 x-2)^{2}+x^{2}$ and the equation to solve is:

$$
10 x^{2}-12 x+3=0
$$

From this, $x=\frac{6 \pm \sqrt{6}}{10}$. The value $x_{1}=\frac{6+\sqrt{6}}{10} \approx 0.8449$ falls within the appropriate interval, while $x_{2}=\frac{6-\sqrt{6}}{10} \approx 0.3551$ does not.

Therefore, the equation has infinitely many solutions: for every integer $n$, there are three in the interval $[n, n+1)$:

$$
n+\frac{1}{\sqrt{10}}, \quad n+\frac{3}{5} \quad \text { and } \quad n+\frac{6+\sqrt{6}}{10}
$$