Given $ab (c + d)+(a + b) cd = 2018$ and $(ab + 1) (cd + 1) + (a + b- 1) (c + d- 1) = -1$ and $a,b,c,d$ are integers, find all solution tuples.
So I have tried to factorize this and get $(a-1)(b-1)(c-1)(d-1)=-2020$ by subtracting second equation from the first; but now I have no idea how to continue. Do I just try all the possible sets of $a,b,c,d$? I think it is a possible way but there are $1280$ possible tuples of $a,b,c,d$; it is just to much to compute. Help will be appreciated!
The problem was changed. In the original problem was the following given.
$$(ab + 1) (cd + 1) + (a + b- 1) (c + d- 1) = 1$$
It should be $$(a-1)(b-1)(c-1)(d-1)=-2018$$ and since $$2018=2\cdot1009,$$ where $1009$ is prime, we have not so many cases.