Problem 11.5. Determine the number of possible values of the product $a \cdot b$, where $a, b-$ are integers satisfying the inequalities

$$
2019^{2} \leqslant a \leqslant b \leqslant 2020^{2}
$$

Answer: $\mathrm{C}_{2 \cdot 2019+2}^{2}+2 \cdot 2019+1=2 \cdot 2019^{2}+5 \cdot 2019+2=8162819$.
Solution. Suppose that for some two different pairs $a \leqslant b$ and $a_{1} \leqslant b_{1}$ from the interval $\left[2019^{2} ; 2020^{2}\right]$, the equality $a b=a_{1} b_{1}$ holds. Without loss of generality, assume that $a_{1} < a$. Then

$$
a-a_{1} \geqslant 2019
$$

and

$$
b-\frac{a_{1}+b_{1}}{2} \geqslant 2020
$$

Adding inequalities (2) and (3), we derive

$$
b-a \geqslant 2019+2020=2020^{2}-2019^{2}
$$

Therefore, $b=2020^{2}$ and $a=2019^{2}$, and all the non-strict inequalities above must turn into equalities. The inequalities of means will turn into equalities only if the terms are equal, that is, $a_{1}=b_{1}=\sqrt{a b}=2019 \cdot 2020$.

## Kurchatov School Olympiad in Mathematics - 2019 Final Round.

7 6. A fully correct solution is provided.

2 6. The uniqueness of the solution to the equation $a b=a_{1} b_{1}$ with different pairs $a \leqslant b$ and $a_{1} \leqslant b_{1}$ is claimed (without proper justification) and the correct answer is obtained.

1 6. Only the number of pairs of different and identical numbers from the interval $\left[2019^{2} ; 2020^{2}\right]$ is found.

Points are deducted for the following flaw in the solution, which is otherwise correct:

-1 6. Arithmetic errors are made that do not affect the course of the solution.