4. Baron Münchhausen claims that there exist two different 10-digit numbers, not divisible by 10, such that if you subtract the sum of the digits of their squares from each of these numbers, the results will be the same. Is the baron lying?
Answer: does not lie.

Solution. Examples of such numbers can be $10^{9}+8$ and $10^{9}+9$. Indeed, the square of the first number is

$$
\left(10^{9}+8\right)^{2}=10^{18}+16 \cdot 10^{9}+64=1 \cdot 10^{18}+1 \cdot 10^{10}+6 \cdot 10^{9}+6 \cdot 10+4
$$

and the sum of its digits is 18. The square of the second number is

$$
\left(10^{9}+9\right)^{2}=10^{18}+18 \cdot 10^{9}+81=1 \cdot 10^{18}+1 \cdot 10^{10}+8 \cdot 10^{9}+8 \cdot 10+1
$$

and the sum of its digits is 19. At the same time, the first number is 1 less than the second.