Hong Kong Preliminary Selection Contest

Let $m$ be an integral root of the equation. Then we have $k{m}^2+(4k-2)m+(4k-7)=0$, which can be rewritten as $(m^2+4m+4)k=2m+7$, or $(m+2{)}^{2}k=2m+7$. Hence $m+2$ divides $2m+7$, and so $m+2$ divides $2m+7-2(m+2)=3$ as well. Thus $m$ can only be $-5,-3,-1$ or $1$, which we plug in to $(m+2{)}^{2}k=2m+7$ one by one: When $m=-5$, we get $9k=-3$ which gives no integral solution for $k$. When $m=-3$, we get $k=1$. When $m=-1$, we get $k=5$. When $m=1$, we get $9k=9$ which again gives $k=1$. Therefore, the sum of all possible values of $k$ is $1+5=6$.