66th Czech and Slovak Mathematical Olympiad

Multiplying both sides by $x+y$ yields $$x^2-xy+2y^2=k(x+y).(1)$$ Any solution $(x,y)$ to the original equation is a solution to (1), but (1) can have extra solutions satisfying $x+y=0$, tj. $y=-x$. A pair $(x,-x)$ is a solution to (1) if and only if $x^2+x^2+2x^2=k\cdot 0$, that is $x=0$. Equation (1) therefore has exactly one more solution than the original one and it suffices to prove that equation (1) has an even number of integer solutions if and only if $7\mid k$. We rewrite (1) as a quadratic equation $$x^2-x(y+k)+2y^2-ky=0(2)$$ in $x$. Its discriminant satisfies $$\begin{array}{rl}D(y)=(y+k{)}^2-4(2y^2-ky)& =k^2+6ky-7y^2=(k-y)(k+7y)=\\ & =-7{\left(y-\frac{3}{7}k\right)}^2+\frac{16}{7}{k}^2,\end{array}\text{\hspace{1em}(3)}$$ which is, for every $k$, a quadratic function in $y$ bounded from above. Therefore, for any integer $k$, the discriminant $D(y)$ is non-negative for only finitely many integers $y$ and equation (2) has only finitely many integer solutions $(x,y)$.

If $D(y)>0$ for some integer $y$, the equation (2) has precisely two real solutions that can only be integer simultaneously, for their sum $y+k$ is an integer. For any such $y$ we get an even number of solutions to (2). We see that $D(y)=0$ either for $y=k$ or for $y=-\frac{1}{7}k$. In the first case, equation (2) reduces to $(x-k{)}^2=0$ with double root $x=k$ and the equation (1) has only one solution $(k,k)$ with $y=k$. In the second case, $y$ is integer if and only if $k$ is divisible by 7 and then equation (2) has a double root $x=\frac{3}{7}k$, therefore $(\frac{3}{7}k,-\frac{1}{7}k)$ is the only solution to equation (1) with $y=-\frac{1}{7}k$. Moreover, the two solutions $(k,k)$ and $(\frac{3}{7}k,-\frac{1}{7}k)$ are different as $k\ne 0$. We see that equation (1) has an even number of integer solutions if and only if $k$ is divisible by 7. We conclude.