Compute $a+b+c+d$ where $(ab+c+d =13),\ (bc+d+a = 27),\ (cd+a+b = 30),\ (da+b+c = 17)$

Question

Compute $(a+b+c+d)\ $ if $\ a,b,c,d\ $ satisfy the system of equations below:

$$\begin{cases} ab+c+d = 13\\\\ bc+d+a = 27\\\\ cd+a+b = 30\\\\ da+b+c = 17 \end{cases}$$

.................................................................................My method:

I added all equations to get:

$(ab+bc+cd+da)+2(a+b+c+d)=87$

$(a+c)(b+d)+2(a+c)+2(b+d)=87$

$(a+c)(b+d+2)+2(b+d)=87$

$(a+c+2)(b+d+2)=91$

$(a+c+2)(b+d+2)=13\times 7$

So, $a+c=11$ and $b+d=5$

or, $a+c =5$ and $b+d =11$

Hence, $a+b+c+d = 16$

But I am not sure if this is the correct method since it was not stated that $a,b,c,d$ were positive integers.

I would be thankful if someone provided a solution to this question.

Answer 1

Let $u = a+b+c+d$, the set of equations can be rewritten as

$$ \begin{cases} (a-1)(b-1) + a + b + c + d &= 14\\ (b-1)(c-1) + a + b + c + d &= 28\\ (c-1)(d-1) + a + b + c + d &= 31\\ (d-1)(a-1) + a + b + c + d &= 18 \end{cases} \iff \begin{cases} (a-1)(b-1) &= 14- u\\ (b-1)(c-1) &= 28 -u\\ (c-1)(d-1) &= 31 -u\\ (d-1)(a-1) &= 18 -u \end{cases}$$ This leads to

$$(14-u)(31-u) = (a-1)(b-1)(c-1)(d-1) = (28-u)(18-u)$$ which can be simplified to $u = 70$. So $a + b + c + d = 70$.