6. Tenth-grader Dima came up with two natural numbers (not necessarily different). Then he found the sum and the product of these numbers. It turned out that one of these numbers coincides with the arithmetic mean of the other three. What numbers did Dima come up with? Find all possible answers and prove that others are impossible.
6. Answer. There are three solutions $(1,2),(4,4)$ and $(3,6)$. Solution. Let the numbers Dim invented be $a, b$. If each of these numbers is different from 1, then their product $a b$ is greater than the other three and cannot be their arithmetic mean. Therefore, there are two fundamentally different cases: $3 a=b+(a+b)+a b$ and $3(a+b)=b+a+a b$. The first equation is equivalent to $2(a-b)=a b \Leftrightarrow \frac{1}{b}-\frac{1}{a}=\frac{1}{2}$. Clearly, the only possible case here is $b=1, a=2$. The second equation is equivalent to $2(a+b)=a b \Leftrightarrow \frac{1}{b}+\frac{1}{a}=\frac{1}{2}$. Then one of these fractions is either $1 / 3$ or $1 / 4$, which gives two cases: $b=4, a=4$ or $b=3, a=6$. Finally, if one of these numbers is 1, we have the set $1, x, x, x+1$. The mean in this set can only be the number $x$, which gives $3 x=2 x+2$ and $x=2$. Grading criteria. Correct solution - 7 points. The reasoning scheme is generally correct but contains logical errors leading to the loss of one of the solutions - 4 points. All equations are written, but not solved or solved incorrectly - 1 point. All solutions are provided and verified, but the absence of other solutions is not proven - 2 points. Two solutions are provided and verified - 1 point. In all other cases - 0 points.