10.3. In a bag, there are chips of two colors, blue and green. The probability of drawing two chips of the same color is equal to the probability of drawing two chips of different colors (drawing one chip after another without replacement). It is known that there are more blue chips than green chips, and the total number of chips is more than 2 but less than 50. How many blue chips can there be in the bag? Find all answers.
Answer: $3,6,10,15,21,28$.

Solution. Let the number of blue chips be $n$, and the number of green chips be $-m$. Then $\frac{n m+m n}{(n+m)(n+m-1)}=\frac{n(n-1)+m(m-1)}{(n+m)(n+m-1)}$, or $2 n m=n(n-1)+m(m-1)$, from which $(n-m)^{2}=n+m$. Thus, the total number of chips (denote it as $a$) must be a perfect square, that is, $a$ can take the values $4,9,16,25,36,49$. The number of blue chips can be found from the system $\left\{\begin{array}{c}n+m=a \\ n-m=\sqrt{a}\end{array}\right.$. Solving the system, we get $n=\frac{a+\sqrt{a}}{2}$. Therefore, $n \in\{3,6,10,15,21,28\}$.

Comment. A correct and justified solution - 7 points. Obtaining the equation $2 n m=n(n-1)+m(m-1)$ - 3 points. Finding possible values of the total number of chips - 5 points. Answers found by trial, showing that they satisfy the condition, but not proving that there are no other answers - 3 points. Only part of the answers found by trial - 2 points. Only one answer found by trial - 1 point. For each arithmetic error, 1 point is deducted. Only answers provided without justification - 0 points.