Consider the solutions of the system

$$
\left\{\begin{array}{l}
2019=a+b-c \\
2019=a^{2}+b^{2}-c^{2}
\end{array}\right.
$$

where $a, b$, and $c$ are integers.

a) Find at least one solution of the system.

b) Verify that the number of solutions is finite.
Solution

a) From the first equation, it follows that $c=a+b-2019$. Substituting into the second equation, we get

$$
2019=a^{2}+b^{2}-(a+b-2019)^{2}=-2 a b+4038 a+4038 b-2019^{2}
$$

Thus,

$$
\begin{aligned}
2019-2019^{2} & =-2 a b+4038 a+4038 b-2 \cdot 2019^{2} \\
-2019 \cdot 2018 & =-2(a-2019)(b-2019) \\
2019 \cdot 1009 & =(a-2019)(b-2019)
\end{aligned}
$$

To obtain a solution, it is sufficient that $a-2019=1$ and $b-2019=2019 \cdot 1009$, that is,

$$
(a, b, c)=\left(2020, \frac{2019 \cdot 2020}{2}, \frac{2019 \cdot 2020}{2}+1\right)
$$

b) Due to the previous item, $a-2019$ and $b-2019$ are integer divisors of the number $2019 \cdot 1009$, which has a finite number of integer divisors.

Note: If $x$ is an integer divisor of $2019 \cdot 1009$ and $y=2019 \cdot 1009 / x$, then

$$
(a, b, c)=(x+2019, y+2019, x+y+2019)
$$

is a solution to the system with $a-2019=x$ and $b-2019=y$.