Factoring question involving 4 integers. For all non-negative integer values of $a$, $b$, $c$, $d$ given that $ac+bd+bc+ad=42$ and $c^2-d^2=12$, then determine all possible values of $a+b+c+d$.
This was a question on one of my previous tests and I am not sure how to do it.

$c^2 - d^2 = 12 \iff (c+d)(c-d)=12$. The only integers $c,d$ satysfying this equality are $c=4, d=2$. Putting these values into $ac + bd + bc + ad = 42$ we get that $7a+7b = 42 \iff a+b=6$. That means that there are 5 quadruples (a,b,c,d), namely (1,5,4,2), (2,4,4,2), (3,3,4,2), (4,2,4,2), (5,1,4,2).
