9.3 How many right-angled triangles with integer sides exist, where one of the legs is equal to 2021.
Answer: 4. Solution. Let the hypotenuse of a right triangle be $x$, one of the legs be $-y$, and the other be 2021. Then, by the Pythagorean theorem, $x^{2}-y^{2}=2021^{2}$, i.e., $(x-y) \cdot(x+y)=2021^{2}$. Considering that $x>y$, we have: $x+y>x-y>0$. Since the prime factorization of 2021 is $2021=43 \cdot 47$, the prime factorization of $2021^{2}$ is $2021^{2}=43^{2} \cdot 47^{2}$, and therefore $2021^{2}$ can be represented as the product of two natural numbers in five ways:

$2021^{2}=1 \cdot\left(43^{2} \cdot 47^{2}\right)=43 \cdot\left(43 \cdot 47^{2}\right)=47 \cdot\left(43^{2} \cdot 47\right)=43^{2} \cdot 47^{2}=(43 \cdot 47) \cdot(43 \cdot 47)$.

Since in the product $(x-y) \cdot(x+y)$ the factors are different and the second factor is greater than the first, we have four systems of two equations:

![](https://cdn.mathpix.com/cropped/2024_05_06_64f33e064e30fde10c06g-2.jpg?height=123&width=1073&top_left_y=1092&top_left_x=183)

Solving each of these systems, we obviously get natural $x, y$ (which follows from the oddness of the right-hand sides). Therefore, there exist four right triangles with integer sides, one of which has a leg equal to 2021.