\section*{Exercise 1 - 051011}

Find a two-digit number that is equal to the sum of the number in the tens place and the square of the number in the ones place!

Prove that there is only one such number!
}

Assuming there is such a number \(z=10 a+b\) with natural numbers \(a, b\) and \(0<a<10, b<10\), then the equation

\[
10 a+b=a+b^{2}
\]

must hold. Therefore, \(9 a=b^{2}-b\), so \(a=\frac{b(b-1)}{9}\).

Since \(a\) is a natural number and \(b\) as well as \(b-1\) cannot both be divisible by 3, either \(b\) or \(b-1\) must be divisible by 9. Given \(b<10\) and \(a \neq 0\), \(b-1\) cannot be divisible by 9, so \(b=9\) must be true. Then \(a\) is 8.

Thus, only the number 89 can satisfy the conditions. Since \(89=8+9^2\) holds, 89 indeed meets the conditions of the problem.