Following is defined:
$$z=\frac{cx+dy}{ax+by}$$
I have these four terms:
$$v_1=b+az$$ $$v_2=a-bz$$ $$v_3=d+cz$$ $$v_4=c-dz$$
The requirements are: $x,y\in\mathbb{Z};v_1,v_2,v_3,v_4\in\mathbb{Q}$. Meaning that x & y are whole numbers and the four terms are all rational.
From these conditions I want to prove that a, b, c, and d have to be rational. This might not be true and if so I would like to see a counter example or proof that falsifies the claim and I would really like to know the restrictions on a, b, c, and d that these four terms give rise to. Can they be all real numbers or only specific ones etc.
I have already tried to algebraically manipulate the term to find a more suitable shape but have not succeeded.
$$a-b\frac{cx+dy}{ax+by}=\frac{\frac{a^2}bx+ay-cx-dy}{\frac a b x+y}$$
I have also tried to go about it by considering the general rules for adding and multiplying rational and irrational numbers, like rational * irrational = irrational and such. But this also did not seem to get me anywhere.
Additionally I tried to divide the problem into many multiple cases depending on if a, b, or z were rational or not and could show this claim for the case that z is rational. But ultimately this wasn't an answer either.
A counterexample follows: $$x=0; \qquad y=1; \qquad a= \sqrt2 + 1; \qquad b=1; \qquad c=-3-2\sqrt2; \qquad d=\sqrt2 - 1 $$
This counterexample was built with the idea of simplifying computations. If you put $(x;y;b)=(0;1;1)$, then the expression for $z$ becomes simply $$z=d$$
I noticed that $a,b,c,d$ are all rational if and only if $z$ is rational. Hence I picked (cleverly) an irrational $d$ and built the counterexample.
I leave you the details, in particular you can check on your own that these values indeed provide a counterexample.