## Task 20/83

We are looking for all triples of consecutive even or odd numbers, where the sum of their squares is a (in decimal notation truly) four-digit number with four identical digits.
If the middle number of the triplet is denoted by $a$, then the smaller one is $a-2$ and the larger one is $a+2$. According to the problem statement, it holds that

$$
(a-2)^{2}+a^{2}+(a+2)^{2}=1111 x
$$

with $x \in N, 1 \leq x \leq 9$, thus

$$
3 a^{2}+8=1111 x \quad ; \quad a=\sqrt{370 x-2+\frac{x-2}{3}}
$$

Since $a \in N$, it must be that $x-2=3 k$ with $k \in N$; therefore, only the values $x_{1}=2, x_{2}=5$ and $x_{3}=8$ are possible for $x$. However, only $x_{2}=5$ yields a natural number $a$:

$$
a=\sqrt{370 \cdot 5-1}=\sqrt{1849}=43
$$

Thus, there is only one triplet that meets the requirements of the problem: (41;43;45). The verification confirms the correctness.