How many positive integral solutions exist for: $ab + cd = a + b + c + d $,where $1 \le a \le b \le c \le d$ ?
I need some ideas for how to approach this problem.
The equation can be rewritten as $$(a-1)(b-1)+(c-1)(d-1)=2.$$
Now there are not many possibilities to consider! If the first product is $0$, the second must be $2$, and if the first product is $1$, so is the second.
If $a=1$, then we need to have $(c-1)(d-1)=2$. Since $1\le c\le d$, this forces $c=2$, $d=3$. And $b$ can be $1$ or $2$, giving the solutions $(1,1,2,3)$ and $(1,2,2,3)$.
If $a>1$, we need $a=2$, else the left hand side is too big. That forces $b=c=d=2$, giving the third solution $(2,2,2,2)$.
Comment: Note that in general $ab+pa+qb=(a+q)(b+p) -pq$. This relative of completing the square is occasionally useful.