Hi, does anyone know how to do this problem? Please include motivation. Thanks

Take the vectors $\left(\begin{array}{c}a\\ c\end{array}\right)$ and $\left(\begin{array}{c}b\\ d\end{array}\right)$. The area of the parallelogram spanned between them is $ad-bc=k$. So if we take the parallelogram $(0, 0), (a, c), (a+b, c+d), (b, d)$ and use Pick's theorem, we find that there are $k$ integer points $(x, y)$ in the parallelogram which are not on the edges connected to $(a+b, c+d)$. This is true because we have $I+\frac{B}{2}-1=k$, and we're counting exactly this number of points: all interior points, exactly half the boundary points (those on the two nearer edges to $(0, 0)$) and excluding one of the mistakenly included corners. Now every real point inside the parallelogram can be written as $x_1(a, c)+x_2(b, d)$ with $0\leq x_1, x_2\leq 1$. Excluding the two far edges, the last inequality must be strict. Writing the $k$ integer points we found as such, we have found $k$ pairs $(x_1, x_2)$ with the required conditions. There can be no more, of course, because $(ax_1+bx_2, cx_1+dx_2)$ must be an integer point inside that parallelogram. I'm not sure where the condition $(a, b)=(c, d)=1$ comes in. If anyone has a contradiction when this condition doesn't hold, please post it.

I think it has something to do with the borders: when (a,c)=(b,d)=1, the borders have no lattice points but the vertices EDIT: oops

No, that's true when $(a, c)=(b, d)=1$. I thought that's what the point of the equalities was too at first, but the solution only works if the vectors are $(a, c)$ and $(b, d)$. I'm just not sure where it could come into play. All I can think of is that $(a, b)=(c, d)=1$ means that there's some linear combination of the vectors with a unit coordinate, but I'm not sure why that'd be important. I even worked it out with $a=b=3, c=1, d=2$ and found $ad-bc=3$ pairs just like I should: $(0, 0), \left(\frac{1}{3},\frac{1}{3}\right), \left(\frac{2}{3}, \frac{2}{3}\right)$. I guess it's there to make some other method of solution easier.