Example 4. Let's solve the equation

$$
x^{3}-[x]=3
$$
Solution. Let's transform the given equation

$$
x^{3}=[x]+3
$$

Based on property $1^{0}$ of the "floor" function, we get:

$$
x^{3}=[x+3]
$$

From this, it follows that $x^{3} \in \mathbf{Z}$.

We construct the graphs of the functions $y=x^{3}$ and $y=[x+3]$. A graphical estimate (Fig. 5) shows that the only root lies between the numbers 1 and 2. Additionally, since $x^{3} \in \mathbf{Z}$ and $x^{3}=[x+3]$, $x$ must be the cube root of some integer $b$ not exceeding 8. Testing $b=1,2,3,4$, we find that $b=4$ and

$$
x=\sqrt[3]{4} \in [1; 2]
$$

satisfies the given equation.

## EXERCISES

47. Solve the equations:

a) $[3.5 x]=20$, where $x \in \mathbf{N} ; \quad$ g) $[x]=x^{3}-21$;

b) $\left[x^{2}\right]=2$ 
c) $\left[\frac{8 x+19}{7}\right]=\frac{16(x+1)}{11}$;

d) $\left[3 x^{2}-x\right]=x+1$;

e) $\left[\frac{2 x-1}{3}\right]=\left[\frac{x+1}{2}\right]$;

f) $[x]=\frac{3}{4} x$

g) $\left[\frac{2 x-1}{3}\right]=\left[\frac{2 x+5}{5}\right]$;

h) $\left[\frac{x^{3}-2}{3}\right]=[x]$;
i) $x^{3}-[x]=-3$

j) $\left\{\begin{array}{l}{[x+y-8]=6-x,} \\ {[x-5]+[y-4]=15-x-y ;}\end{array}\right.$

k) $\sqrt{\left[-7 x^{2}+3 x+4\right]}=[2-\sin x]$.

48*. Solve the inequality

$$
\left|x-\left[x+\frac{1}{2}\right]\right| \leqslant \frac{1}{2}
$$

49*. Determine the number of non-negative integer solutions to the equation

$$
\left[\frac{x}{a}\right]=\left[\frac{x}{a+k}\right]
$$

where $a, k \in \mathbf{N}$.

50. Show that the solution to the equation

$$
[x]+\left[x+\frac{1}{100}\right]+\left[x+\frac{2}{100}\right]+\ldots+\left[x+\frac{99}{100}\right]=[100 x]
$$

is any $x \in \mathbf{R}$.

51*. Find the number of all natural numbers ending in at least two identical, non-zero digits and not exceeding a given number $N$.

52. A traveler was on the road for an integer number of days, traveling as many kilometers each day as the number of days he was on the road. If he had traveled 20 km each day and stopped for one day after every 40 km, his travel time would have increased by 37 days. How many days was the traveler on the road?

## 26

## § 4. Functional Equations

A functional equation is a relation expressing a certain property possessed by a class of functions (a specific function).

Simple examples of functional equations include: $f(x)=f(-x)$ - the equation of evenness, $f(x+T)=f(x)$ - the equation of periodicity, $f(x+y)=f(x)+f(y)$, and others.

A function $f(x)$ is called a solution to a given functional equation if it satisfies the equation for all values of the argument in its domain.

For example, the functions

$$
f(x)=a x^{2}, f(x)=\sin 2 \pi x, f(x)=a x
$$

where $a \in \mathbf{R}$, are particular solutions to the respective equations above, as can be easily verified by substitution.

To solve a functional equation means to determine whether it has solutions and to find them if they exist.

The process of finding a solution is determined by both the set goal and the restrictive conditions expressing certain properties that the sought function must possess (monotonicity, continuity, differentiability, etc.). For example, the equation

$$
f(x+y)=f(x)+f(y)
$$

under the assumption of the continuity of the sought function has a unique solution $f(x)=a x$, but if this condition is omitted, a discontinuous function can be indicated that satisfies this equation.

We will mention some methods for solving functional equations:

1. Cauchy's Method. This method involves finding the solution to a functional equation step by step for all natural, integer, rational, and then real values of the argument. Clearly, this method assumes the continuity of the functions.

Example 1. Solve the equation

$$
f(x+y)=f(x) \cdot f(y)
$$

Solution. Setting $y=x, 2x, \ldots$, we get (by induction):

$$
\begin{aligned}
& f(2 x)=f^{2}(x) \\
& f(3 x)=f^{3}(x) \\
& \dot{f}(n x)=f^{n}(x)
\end{aligned}
$$

Taking $x=1$ in these equations, we get:

$$
f(n)=f^{n}(1)
$$

Assuming $f(1)=C_{1}$, we have for $x \in \mathbf{N}: f(x)=C_{1}^{x}$.

Setting $x=\frac{m}{n}$ in equation (2), we get:

$$
f(m)=f^{n}\left(\frac{m}{n}\right)=C_{2}^{m}
$$

from which

$$
f\left(\frac{m}{n}\right)=C_{2}^{\frac{m}{n}}
$$

Thus, for $x \in \mathbf{Q}$, the solution to the given equation is $f(x)=C^{x}$. Assuming that the function $f(x)$ is continuous, we get: $f(x)=C^{x}$ for $x \in \mathrm{R}, x>0$.

Setting $x=0$ in equation (1), we have:

$$
f(y)=f(0) \cdot f(y)
$$

from which $f(y)=0$, or $f(0)=1=C^{0}$.

Setting $y=-x$ in equation (1), we find:

$$
\begin{gathered}
f(0)=f(-x) f(x), \quad 1=f(-x) f(x) \\
f(-x)=\frac{1}{f(x)}=\frac{1}{C^{x}}=C^{-x}
\end{gathered}
$$

Thus, for any real $x$, the solution to the equation is the function $f(x)=C^{x}$.

2. Substitution Method. This method involves substituting different values for $x$ (or $y$) and combining the resulting equations with the original one to obtain (usually by elimination) an algebraic equation in terms of the sought function.