## Task 35/80

We are looking for all four-digit natural numbers $n$ with the following properties:

1. The square of the second digit is equal to the number $n_{1}$, which is obtained by removing the first two digits from $n$.
2. The sum of the digits $Q(n)$ is equal to the number $n_{2}$, which is obtained by removing the last two digits from $n$.
3. The sum of the numbers $n_{1}$ and $n_{2}$ is 100.
Let $n=1000 a+100 b+10 c+d$ with $a, b, c, d \in \mathbb{N}, 0 \leq a, b, c, d \leq 9$. Then $n_{1}=10 c+d$ and $n_{2}=10 a+b$, and it holds that

according to condition 1: $b^{2}=10 c+d \quad(1)$

according to condition 2: $a+b+c+d=10 a+b$ thus $c+d=9 a$

according to condition 3: $10 a+b+10 c+d=100$

Since $c+d \leq 18$, it immediately follows from (2) that $a=1$ or $a=2$. However, the possibility $a=2$ is ruled out as it requires $c=d=9$, which contradicts (1) ($b^{2}=99$ is not satisfied for any natural number $b$). Therefore, $a=1$, and the relationships (2) and (3) take the form:

$$
c+d=9 \quad\left(2^{\prime}\right) \quad ; \quad b+9 c=81
$$

Thus, either $b=0, c=9, d=0$ or $b=9, c=8, d=1$. The first possibility again contradicts (1). Therefore, there exists exactly one number $n$ with the required properties: $n=1981$. The verification confirms the correctness.