\section*{Problem 2 - 111032}

Determine all ordered pairs \((x ; y)\) of two-digit natural numbers \(x\) and \(y\) with \(x>y\) for which the following holds:

a) If the digits of the number \(x\) are written in reverse order, the number \(y\) is obtained.

b) If the digits of the number \(x^{2}\) are written in reverse order, the number \(y^{2}\) is obtained.
Assume there is a pair of numbers \((x, y)\) that satisfies conditions a) and b). If we set \(x=10a+b\) (with \(a, b\) natural and \(1 \leq a \leq 9, 1 \leq b \leq 9\)), then it follows that \(y=10b+a\), and because \(x>y\), we also have \(a>b\).

Since \(a \neq 0, b \neq 0\), and \(a > b\), it follows that \(2 \leq a \leq 9\) and \(1 \leq b \leq 8\).

The square of the two-digit number \(x\) is either three-digit or four-digit. We first consider the cases where \(x^2\) is three-digit. Since \(40^2 > 1000\), it follows that \(a \leq 3\).

Since \(32^2 = 1024\) is already four-digit, the numbers 21 and 31 are the only candidates that could satisfy the conditions. Indeed,

\[
21^2 = 441 \quad \text{and} \quad 12^2 = 144 \quad \text{as well as} \quad 31^2 = 961 \quad \text{and} \quad 13^2 = 169
\]

Thus, the pairs \((21, 12)\) and \((31, 13)\) satisfy conditions a) and b).

Now assume that the conditions of the problem can be satisfied by a number \(x\) whose square \(x^2\) is four-digit. Then for the digits \(a, b\) of this number,

\[
\text{(3)} \quad (10a + b)^2 = 1000c + 100d + 10e + f \quad \text{and} \quad \text{(4)} \quad (10b + a)^2 = 1000f + 100e + 10d + c
\]

(with \(c, d, e, f\) natural and \(0 \leq c, d, e, f \leq 9; c, f \neq 0\)). From (3) and (4) it follows that

\[
100a^2 + 20ab + b^2 = 1000c + 100d + 10e + f \quad ; \quad a^2 + 20ab + b^2 = c + 10d + 100e + 1000f
\]

By subtraction, we obtain

\[
99a^2 - 99b^2 = 999c + 90d - 90e - 999f \quad \text{thus}
\]

(5) \(11(a^2 - b^2) = 111c + 10d - 10e - 111f\)

Since the left side of (5) is divisible by 11, the right side must also be divisible by 11.

If we add to \(111c + 10d - 10e - 111f\) the number \(1111f + 110e - 110c\) (which is divisible by 11), we get \(1000f + 100e + 10d + c = (10b + a)^2\), and this number must also be divisible by 11.

Therefore, \(11 \mid (10b + a)\) must hold, which is not the case because \(a \neq b\). This contradiction proves that there is no such pair \((x, y)\) for four-digit \(x^2\).

\section*{Adapted from \([5]\)}