Find all integer tuples $(a,b,c,d)$ such that $$ab-2cd-3=ac+bd-1=0.$$
The substitution $ab=p,cd=q,ac=r$ and $bd=s$, leads to a quadratic in $r$ (messy algebra)
$$r^2-r+(3q+2q^2)=0$$
Since the discriminant of the above quadratic is non-negative, $1-12q-8q^2>0$ which implies $(q,r)=(0,0)$ or $(q,r)=(0,1)$.
Tedious casework leads to
$$(a,b,c,d)\in\{(1,3,1,0),~(-1,-3,-1,0),~(3,1,0,1),~(-3,-1,0,-1)\}$$
I am looking for a solution using factorization or something more clean. Adding the two equations seems promising, but I just can't make it work.
Rewrite the two given equations as $$ ab-2cd=3 \quad\text{and}\quad ac+bd=1, $$ and note for later use that $a\ne0$ and $b\ne0$ and that at most one $c$ and $d$ can equal $0$, and also that at least one variable is even. Squaring both sides of both equations and rearranging yields $$ a^2b^2+4c^2d^2=9+4abcd \quad\text{and}\quad a^2c^2+b^2d^2=1-abcd, $$ and then adding the first equation to $4$ times the second equation gives $$ a^2b^2+4c^2d^2+4a^2c^2+4b^2d^2=13. $$ From here we conclude that exactly one variable equals $0$ (if none equals $0$, then the left-hand side is greater than $13$ unless $|a|=|b|=|c|=|d|=1$, contradicting the fact that one must be even). If $c=0$ then the equation reduces to $$ b^2(a^2+4d^2)=13 $$ which implies $|b|=1$ and $|d|=1$ and thus $|a|=3$, and the proper signs are easy to work out; if $d=0$ then the argument is the same with the variables permuted.