We want
$$\left(a+\frac{b}{c}\right)\left(x-\frac{y}{z}\right)=a x$$
which means
$$b y+a c y-b x z=0$$
which we can solve for $x$ as
$$x=\frac{b y+ a c y}{b z}$$
The task now becomes finding set of five-tuples $(a,b,c,y,z)$ such that $\frac{b y+ a c y}{b z}$ is an integer number and we have the constraints $b\le c-1$ and $y\le z-1$ as any integer part of the fractions $b/c$ and $y/z$ can be absorbed into $a$ and $x$.
For $a,b,c,y,z=1,\dots,20$ we do have 32961 proper solutions, which my 2015 Macbook with 2 cores finds in 2.3 seconds. The code is
results = With[{k = 20},
Select[IntegerQ[#[[4]]] &]@
Flatten[Table[{a, b, c, (b y + a c y)/(b z), y, z}, {a, k}, {c,
k}, {b, c - 1}, {z, k}, {y, z - 1}], 4]
]; // RepeatedTiming
(* {2.3, Null} *)
As examples:
HoldForm[(a + b/c) (x - y/z)] /. Table[
Thread[{a, b, c, x, y, z} -> results[[i]]], {i, {456, 4956, 32961}}
]
(* {(1+1/11) (6-4/8),(4+2/6) (12-12/13),(20+18/20) (22-18/19)} *)
where the results are integers as designed:
% // ReleaseHold
(* {6, 48, 440} *)
One can technically increase the efficiency by noting that the pairs $(b,c)$ and $(y,z)$ are defined only upto an overall scaling, i.e. $(x,y)\sim (\lambda x,\lambda y)$ for integer $\lambda$. Hence we can restrict the search for integers $(b,c)$ which are coprime, and similarly for $(y,z)$. In other words, we can first generate all coprimes within a certain range and then use them instead. For the integers in the range $1\cdots 30$, we run
results = With[{k = 30},
With[{
pairs =
Select[CoprimeQ @@ # &]@
Flatten[Table[{j, i}, {i, k}, {j, i - 1}], 1]
},
Select[IntegerQ[#[[4]]] &]@Flatten[Table[
With[{b = p1[[1]], c = p1[[2]], y = p2[[1]], z = p2[[2]]},
{a, b, c, (b y + a c y)/(b z), y, z}
], {a, k}, {p1, pairs}, {p2, pairs}], 2]
]]; // RepeatedTiming
(* {35., Null} *)
results // Length
(* 34406 *)
As an example, we see that
HoldForm[(a + b/c) (x - y/z)] /. Thread[{a, b, c, x, y, z} -> results[[-1]]]
(* (30+1/30) (848-16/17) *)
which is
% // ReleaseHold
(* 25440 *)
One can also generate examples of larger numbers using FindInstance command, as suggested by others. In terms of our equations, it reads as
FindInstance[b y + a c y - b x z == 0 && c > b && z > y
&& a > 10000, {a, b, c, x, y, z}, PositiveIntegers, 3]
(* {{a -> 10247, b -> 1, c -> 42, x -> 11, y -> 1, z -> 39125}, {a -> 10247, b -> 1, c -> 42, x -> 11, y -> 71, z -> 2777875}, {a -> 11414, b -> 1, c -> 23, x -> 2, y -> 44, z -> 5775506}} *)
which gives
HoldForm[(a + b/c) (x - y/z)] /. %
(* {(10247+1/42) (11-1/39125),(10247+1/42) (11-71/2777875),(11414+1/23) (2-44/5775506)} *)
For example,
$$\left(11414+\frac{1}{23}\right)\left(2-\frac{44}{5775506}\right)=11414\times 2$$
