Problem 3. On the board, the numbers $2,3,5, \ldots, 2003,2011,2017$ are written, i.e., all prime numbers not exceeding 2020. In one operation, two numbers $a, b$ can be replaced by the largest prime number not exceeding $\sqrt{a^{2}-a b+b^{2}}$. After several operations, only one number remains on the board. What is the maximum value it can take?
Answer: The maximum value is 2011.

Solution. Note that if $a$ $2011^{2}$, therefore, as a result of this operation, the number 2011 will appear on the board.

Remark. The inequality $a<\sqrt{a^{2}-a b+b^{2}}<b$ for $a<b$ can be proven geometrically. The matter is that in a triangle with sides $a$ and $b$ and an angle of $60^{\circ}$ between them, the length of the third side is exactly $\sqrt{a^{2}-a b+b^{2}}$. Since the angle $60^{\circ}$ is always the middle in size in its triangle, the side opposite to it is also the middle in size.

## Criteria

Any complete solution to the problem is worth 7 points. In the absence of such a solution, the following criteria are summed up:

3 p. It is shown how to get 2011, but it is not proven that a larger number cannot be obtained.

4 p. It is proven that a number larger than 2011 cannot be obtained.