Problem 7.2. For natural numbers $m$ and $n$, it is known that if their sum is multiplied by the absolute value of their difference, the result is 2021. What values can the numbers $m$ and $n$ take? (7 points)

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# Solution.

Let $\mathrm{m} \geq \mathrm{n}$, then the modulus of their difference is (m-n). Since $\mathrm{m}$ and $\mathrm{n}$ are natural numbers, their sum and the modulus of their difference are also natural numbers. The number 2021 is divisible only by $1,43,47$

and 2021. Therefore, there are only four possible cases: $\left\{\begin{array}{c}m-n=1 \\ m+n=2021\end{array} ;\left\{\begin{array}{c}m-n=43 \\ m+n=47\end{array} ;\left\{\begin{array}{c}m-n=2021 \\ m+n=1\end{array}\right.\right.\right.$; $\left\{\begin{array}{l}m-n=47 \\ m+n=43\end{array}\right.$ The third and fourth cases can be disregarded given the statement that the sum of natural numbers is greater than the modulus of their difference. Solving the systems of equations, we get two answers: $\mathrm{m}=1011, \mathrm{n}=1010$ and $\mathrm{m}=45, \mathrm{n}=2$.

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