<s>[INST] Compute $a+b+c+d$ where $(ab+c+d =13),\ (bc+d+a = 27),\ (cd+a+b = 30),\ (da+b+c = 17)$ 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.
 [/INST] 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 $a + b + c + d = 70$.
</s>