# Problem 2. (Folklore)

In a box, there are balls of two colors: blue and red (both colors are present). It is known that there are more blue balls, and two balls of the same color can be drawn with the same probability as two balls of different colors. What can the difference between the number of blue and red balls be? Provide a complete and justified answer to this question.
Answer: Any natural number greater than 1. Solution:

Let blue be $m$, red be $n(m>n)$.

The probability of drawing two balls of the same color $\left(\frac{n}{n+m}\right)\left(\frac{n-1}{n+m-1}\right)+\left(\frac{m}{n+m}\right)\left(\frac{m-1}{n+m-1}\right)$

The probability of drawing two balls of different colors $\left(\frac{n}{n+m}\right)\left(\frac{m}{n+m-1}\right)+\left(\frac{m}{n+m}\right)\left(\frac{n}{n+m-1}\right)$

That is, $n(n-1)+m(m-1)=n m+m n \Rightarrow(m-n)^{2}=m+n$

Then, denoting $m-n=a, \quad a^{2}=n+n+a \Rightarrow a^{2}-2 n-a=0 \Rightarrow n=\frac{a^{2}-a}{2}, m=\frac{a^{2}+a}{2}$

$a=1$ does not work (since $n>0$ by condition) For $a>1 m$ and $n$ are natural (since $a$ and $a^{2}$ have the same parity)

Criteria: Correct solution: 15 points (+)

Formulas for probabilities do not need explanation, but if they are significantly incorrect (for example, if two balls of the same color are taken unordered, and different ones are ordered) -0 points $(-)$

Without explanation, if $n(n-1)+m(m-1)=n m+m n$ and so on is written directly: no more than 11 points ( $\pm$ )

If the expression $n=\frac{a^{2}-a}{2}$ is found in the solution, it is necessary to understand when $n(a)$ is natural.

For this, it is necessary to say that $n(a)$ is an integer, and that $n(a)>0 \Leftrightarrow a>1$

If only one of these is checked: 13 points (+.), nothing: 11 points ( $\pm$ )

Expressed $m+n($ or $\sqrt{m+n})$ through the difference - no less than 8 points $\left({ }^{+} / 2\right), m$ or $n-$ no less than 11 ( $\pm$ ) Expressed the difference through $m$, and wrote the formula $k(m)$ in the answer - if $k(m)$ is always an integer, 11 points ( $\pm$ )