$\left[\begin{array}{ll}\text { Decimal numeral system } \\ \text { [Examples and counterexamples. Constructions] }\end{array}\right]$

Do there exist three non-zero digits that can be used to form an infinite number of decimal representations of squares of different integers?
Answer: Yes, they do exist. For example, the digits 1, 5, 6 fit. Indeed, let's prove that $\underbrace{3 \ldots 3}_{n \text { digits }} 4^{2}=$ $\underbrace{1 \ldots 1}_{n+1} \underbrace{5 \ldots 5}_{n} 6$. For this, note that $\underbrace{3 \ldots 3}_{n \text { digits }} 5 \cdot 3=\underbrace{10 \ldots 0}_{n}$. Using the identity $n^{2}=1+(n-1)(n+$ 1), we get

$$
\underbrace{3 \ldots 3}_{n \text { digits }} 4^{2}=1+\underbrace{3 \ldots 3}_{n+1} \cdot \underbrace{3 \ldots 3}_{n \text { digits }} 5=1+\underbrace{1 \ldots 1}_{n+1} \cdot 3 \cdot \underbrace{3 \ldots 3}_{n \text { digits }} 5=1+\underbrace{1 \ldots 1}_{n+1} \cdot 1 \underbrace{0 \ldots 0}_{n} 5=\underbrace{1 \ldots 1}_{n+1} \underbrace{5 \ldots 5}_{n} 6
$$

## [ Irrational Equations $\quad]$ Problem 108984 Topics: [Completing the Square. Sums of Squares] [ Substitution of Variables (Other). $\quad]$

Find all real solutions of the equation

$$
36 / \sqrt{x-2}+4 / \sqrt{y-1}=28-4 \sqrt{x-2}-\sqrt{y-1}
$$

## Solution

From the condition, it is clear that $x \geq 2, y \geq 1$. Let's introduce the notations: $\sqrt{x-2}=a^{2}, \sqrt{y-1}=b^{2}$. Our equation can be rewritten as: $36 / a^{2}+4 / b^{2}+4 a^{2}+b^{2}-28=0$. We will split the constant term into two addends and regroup the equations:

$$
\left(36 / a^{2}-24+4 a^{2}\right)+\left(4 / b^{2}-4+b^{2}\right)=0
$$

or

$$
(6 / a-2 a)^{2}+(2 / b-b)^{2}=0
$$

But the sum of two squares of real numbers can equal zero only if each addend is zero. Therefore, $6 / a=2 a, 2 / b=b$ or $a^{2}=3, b^{2}=2$. Returning to the original notations, we find: $x=11, y=5$.

## Answer

$x=11, y=5$.