$ x^{2} -( k+y) x+2y^{2} -ky=0\cdots ( \heartsuit )$ , $ x+y\neq 0$ Note that $ \begin{cases} x+y=0\ \text{and} \ ( \heartsuit ) \ \text{if and only if} \ x=y=0\\ ( x,y) =( k,0) \ \text{is solution of} \ ( \heartsuit )\\ ( x,y) =( k,k) \ \text{is solution of} \ ( \heartsuit ) \end{cases}$ . Since $ x+y\neq 0$, $ ( x,y) =( 0,0)$ is not solution of $ ( \heartsuit )$. Suppose that $ ( x,y) =( \alpha ,\beta )$ is a solution and $ ( \alpha ,\beta ) \neq ( k,0) ,( k,k)$. Then $ ( x,y) =( \beta -\alpha +k,\beta )$ is also solution of $ ( \heartsuit )$ and $ ( \beta -\alpha +k,\beta ) \neq ( k,0) ,( k,k)$. Suppose that the number solutions of $ ( \heartsuit )$ is odd. There exist $ ( \alpha ,\beta ) \in \mathbb{Z}^{2}$ such that $ ( x,y) =( \alpha ,\beta )$ is a solution, $ ( \alpha ,\beta ) \neq ( k,0) ,( k,k) \cdots ( \spadesuit )$, and $ \beta -\alpha +k=\alpha \cdots ( \clubsuit )$ ; otherwise the number of solution is even. Since $ ( x,y) =( \alpha ,\beta )$ is a solution, we have $ \alpha ^{2} -( k+\beta ) \alpha +2\beta ^{2} -k\beta =0\cdots ( \diamondsuit )$. From $ ( \clubsuit )$ and $ ( \diamondsuit )$, we have $ ( k+3\beta )^{2} =16\beta ^{2}$, implying $ k\in \{-7\beta ,\beta \}$. Assume that $ k=\beta $. From $ ( \clubsuit )$, we have $ \alpha =k$. So we have $ \alpha =\beta =k$, which contradicts $ ( \spadesuit )$. So we must have $ k=-7\beta $. Hence we have $ 7\mid k$ . $ \blacksquare $