You are supposed to solve this (part 1)

We have:

$\begin{cases} ab + ac + bc + d = 2 \\ ab + ad + bd + c = 2 \\ ac + ad + cd + b = 2 \\ bc + bd + cd + a = 2 \end{cases}$

Let turn this into the system with 6 equations (hint: use subtraction):

$\begin{cases} (a - b)(c + d - 1) = 0 (1) \\ (a - c)(b + d - 1) = 0 (2) \\ (a - d)(b + c - 1) = 0 (3) \\ (b - c)(a + d - 1) = 0 (4) \\ (b - d)(a + c - 1) = 0 (5) \\ (c - d)(a + b - 1) = 0 (6) \end{cases}$

We will divide this into 4 cases.

Case 1: $a = b = c = d$ then from the first equation of the problem we have $3a^2 + a = 2$ .

Solving this yields 2 solutions: $a = b = c = d = -1$ or $a = b = c = d = \frac{2}{3}$ .

Case 2: $a = b = c \neq d$ then from (6) we have: $a + b = 1$ which also means $2a = 1$ hence $a = b = c = \frac{1}{2}$ .

Moreover, from (1) we have $3a^2 + d = 2$ hence $d = \frac{5}{4}$ , counting permutations which gives us 4 more solutions.

Case 3: $a = b \neq c$ and $a \neq d$ from (2), (3), (4) and (5) we have: $b + d = b + c = a + d = a + c = 1$ that mean $c = d$ . Moreover, from the first equation of the problem we also have $c = 1 - d$ plug it back gives us $a^2 - a + 1 = 0$ which has no real solutions.

Case 4: None of $a$ , $b$ , $c$ , $d$ are the same. From (1) and (2) we can conclude that: $c + d = b + d = 1$ which means $c = b$ . Hence, contradiction.

We have $2 + 4 = \boxed{6}$ solutions in total.