# 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.

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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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Thanks for the hint,I can see only two solution $(1,2,2,3)$ and $(2,2,2,2)$ (till now).However I think there is not any more?! – VelvetThunder Aug 30 '11 at 19:08
Just found and edited almost at the same time you made your comment! Thank you :-) – VelvetThunder Aug 30 '11 at 19:14
Thanks just noticed that too! and when I refreshed you have edited! – VelvetThunder Aug 30 '11 at 19:17
I guess that might be a good gesture for future readers. – VelvetThunder Aug 30 '11 at 19:29

Hint: start by comparing $ab$ to $a+b$.

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