Italian Mathematical Olympiad 1989 - Problem 1

Problem:
Decide if the equation $x^2+xy+y^2=2$ has integer solutions $(x,y)$ where $x$ e $y$ are both rational numbers. 

Solution:
Suppose that the equation $x^2+xy+y^2=2$ has at least one rational solution $\left(\dfrac{x_1}{x_2}, \dfrac{y_1}{y_2} \right)$, where we can suppose $(x_1,x_2), (y_1, y_2) \in \mathbb{Z} \times \mathbb{Z^\ast}$, with $x_1, x_2$ relatively primes and $y_1, y_2$ relatively primes.
We get $$\left(\frac{x_1}{x_2}\right)^2+\frac{x_1}{x_2}\frac{y_1}{y_2}+\left(\frac{y_1}{y_2}\right)^2=2 \Longleftrightarrow x^2_1y^2_2 + x_1x_2y_1y_2 + x^2_2y^2_1 = 2x^2_2y^2_2.$$
Setting $a=x_1y_2, b=x_2y_1, c=x_2y_2$, we obtain the equation $$a^2+ab+b^2=2c^2.$$ Suppose that this equation has an integer solution $(a, b, c)$. If this one is a solution, then also $(-a, -b, -c)$ is a solution. Since $c \neq 0$, we can assume  $c > 0$ and choose $\max (a, b, c)>0$ as small as possible. Now, it is clear that $a$ and $b$ are both even, otherwise we have $1 \equiv 0 \pmod{2}$ and also $c$ is even, otherwise we have $0 \equiv 2 \pmod{4}$, contradiction. So, $a=2a_0, b=2b_0, c=2c_0$, $a_0,b_0,c_0 \in \mathbb{N}, c_0 > 0$, therefore we have $$4a^2_0+4a_0b_0+4b^2_0=8c^2_0 \iff a^2_0+a_0b_0+b^2_0=2c^2_0.$$ Then $(a_0, b_0, c_0)$ is a solution of the equation and $\max(a_0,b_0,c_0) < \max (a, b, c)$ which contradicts the minimality of $\max (a, b, c)$.