I am trying to find out if there are odd positive integers $a,b,c,d$ such that $a,c \equiv 1(4)$ and $b,d \equiv 3(4)$ and $(ab-cd)|(a+b-c-d)$
I tried to prove they are coprime by trying to guess a solution for $x(ab-cd)+y(a+b-c-d) = 1$ but no success.
Thanks for any help
EDIT: also special interest in $a,b,c,d$ greater than 1 and pairwise not equal
Write $a=4e+1,\,b=4f-1,\,c=4g+1,\,d=4h-1$ so $ab-cd=4(4(ef-gh)+f-e+g-h),\,a+b-c-d=4(e+f-g-h)$. The problem is thus equivalent to $4(ef-gh)+f-e+g-h|e+f-g-h$. In the special case $e=g,\,f=h$ we get the vacuous truth $0|0$. This gives the (fairly obvious but uninteresting) solution $a=c,\,b=d$.
There are plenty of other solutions. This code finds solutions with $0<a,\,b,\,c,\,d<100$, and for each prints $a,\,b,\,c,\,d,\,a+b-c-d,\,ab-cd$. An example is $a=1,\,b=19,\,c=5,\,d=3$, viz. $4|12$.