
7. Find all integers $a, b, c, d$ satisfying the following relations:

(i) $1 \leq a \leq b \leq c \leq d$;

(ii) $a b+c d=a+b+c+d+3$.


Solution: We may write (ii) in the form

$$
a b-a-b+1+c d-c-d+1=5
$$

Thus we obtain the equation $(a-1)(b-1)+(c-1)(d-1)=5$. If $a-1 \geq 2$, then (i) shows that $b-1 \geq 2, c-1 \geq 2$ and $d-1 \geq 2$ so that $(a-1)(b-1)+(c-1)(d-1) \geq 8$. It follows that $a-1=0$ or 1 .

If $a-1=0$, then the contribution from $(a-1)(b-1)$ to the sum is zero for any choice of $b$. But then $(c-1)(d-1)=5$ implies that $c-1=1$ and $d-1=5$ by (i). Again (i) shows that $b-1=0$ or 1 since $b \leq c$. Taking $b-1=0, c-1=1$ and $d-1=5$ we get the solution $(a, b, c, d)=(1,1,2,6)$. Similarly, $b-1=1, c-1=1$ and $d-1=5$ gives $(a, b, c, d)=(1,2,2,6)$. In the other case $a-1=1$, we see that $b-1=2$ is not possible for then $c-1 \geq 2$ and $d-1 \geq 2$. Thus $b-1=1$ and this gives $(c-1)(d-1)=4$. It follows that $c-1=1$, $d-1=4$ or $c-1=2, d-1=2$. Considering each of these, we get two more solutions: $(a, b, c, d)=(2,2,2,5),(2,2,3,3)$.

It is easy to verify all these four quadruples are indeed solutions to our problem.

