<s>[INST] Number of positive integral solutions for $ab + cd = a + b + c + d $ with $1 \le a \le b \le c \le d$ 
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.
 [/INST] 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.
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