8.2. Two cyclists, Andrey and Boris, are riding at a constant and identical speed along a straight highway in the same direction, so that the distance between them remains constant. There is a turnoff to a village ahead. At some point in time, the distance from Andrey to the turnoff was equal to the square of the distance from Boris to the same turnoff. When each of them had traveled another 1 km, the distance from Andrey to the turnoff became three times the distance from Boris to the turnoff. What is the distance between the cyclists?
Answer: 2 or 0 km.

Solution: Let the first mentioned distances be $a$ and $b$, then $a=b^{2}$. When each of them had traveled another kilometer, the remaining distance to the turn was $a-1$ and $b-1$ km, respectively, so $a-1=3(b-1)$, which means $b^{2}-1=3(b-1),(b-1)(b+1)=3(b-1)$, from which $b=1$ or $b=2$. When $b=1$, we have $a=1$, so $a-b=0$. When $b=2$, we have $a=4$, so $a-b=2$.

Comment: A complete and justified solution - 7 points. Both equations are correctly set up, but the answer is not obtained - 3 points. The answer is found by selecting numbers that satisfy the condition, but it is not shown that other answers are impossible - 2 points, if only part of the answer is found by selection - 1 point. The solution is correctly started, but there is no significant progress - 1 point. Errors in transformations are made in an otherwise correct solution - deduct 2 points per error. Only the answer is provided - 0 points. The problem is not solved or is solved incorrectly - 0 points.