Ordered quadruples of complex numbers

Adding all of the $4$ equations, we get $ab+bc+cd+ad=0$ , so $(a+c)(b+d)=0$ . Hence either $a+c=0$ or $b+d=0$ .

Case $1$ . $a+c=0$ .

We now know that $a=-c$ . We now substitute. The 4 equations become $-bc=b-c$ , $bc=c-d$ , $cd=d-(-c)$ --> $cd=c+d$ , and $-cd=-c-b$ --> $cd=c+b$ . Take note of the last 2 equations: $cd=c+d$ and $cd=c+b$ . Thus $c+d=c+b$ , and $b=d$ .

Now we substitute $d$ for $b$ to these 4 equations: $-bc=b-c$ , $bc=c-d$ , $cd=c+d$ , and $cd=c+b$ . We get $-dc=d-c$ , $dc=c-d$ , $cd=c+d$ , and $cd=c+d$ . The first two and last two are identical, so we just take the 2nd and the 3rd.

We have $cd=c-d$ and $cd=c+d$ . Hence $c-d=c+d$ and $d=0$ . Thus $c(0)=c-0$ --> $c=0$ . We could also get that $a=b=0$ because $a=-c$ and $b=d$ .

Hence, we only have $(a, b, c, d)=(0, 0, 0, 0)$ .

Case $2$ . $b+d=0$ .

Actually, this case is identical to the first because we will just be shifting the values of $(a, b, c, d)$ . one time to the right, so $(a, b, c, d)$ becomes $(b, c, d, a)$ . Since the first case only has $(0, 0, 0, 0)$ , the second also would only have $(0, 0, 0, 0)$ .

Thus, there is only $\boxed{1}$ such ordered quadruple of complex numbers.

Adding the first two equations together gives $b(a+c) = b-d$ Doing this for the other 3 variables gives also: $a(b+d) = a-c$ $c(b+d) = c-a$ $d(a+c) = d-b$

From equations 2 and 3 we get $a = -c$ , and from 1 and 4 we get $b = -d$ .

Substituting these results into the left-hand side we get $0 = a-c$ and $0 = b-d$ .

Therefore $a=b=c=d=0$ and there is only $\boxed{1}$ solution.

For easy reference, we number the four equations (1), (2), (3), (4), in that order.

Adding the four equations together gives us $ab+bc+cd+ad=0$ , or $(a+c)(b+d)=0$ This implies $a=-c$ or $b=-d$ , so we consider two cases:

Case 1 : $a=-c$ .

Substituting this into equation (1), we get $-bc=b-c$ , or $bc=c-b$ . But from equation (2) we know that $bc=c-d$ , therefore $b=d$ .

Substituting $d=b$ and $a=-c$ into equation (3), we get $bc=b+c$ . We also know from above that $bc=c-b$ , so adding the two equations together yields $2bc=2c$

Therefore $c=0$ or $b=1$ . If $c=0$ then $a=-c=0$ and from equation (1), $0=b-0$ so $b=d=0$ . We get one solution $(0,0,0,0)$

On the other hand, if $b=1$ then $d=b=1$ and since $a=-c$ , equation (1) would give us $-c=1-c$ , which is not possible. Hence case 1 only gives us one solution $(0,0,0,0)$

Case 2 : $b=-d$ .

This is solved very similarly to case 1. We would arrive at $a=c$ and by the symmetry of the equations, we would also get only one solution $(0,0,0,0)$

Thus there is only $\fbox{1}$ quadruple of complex numbers satisfying the four equations.

Adding the four equations gives us $ab+bc+cd+ad=0$

Furthermore, this can be rewritten: $ab+bc+cd+ad=a(b+d)+c(b+d)=(a+c)(b+d)=0$

So either $a=-c$ and/or $b=-d$ .

Starting with substituting the first into our original equations we get;

A: $ab=b+a$ , $-ab=-a-d$ , $-ad=d-a$ and $ad=a-b$

Adding the first two of these we get $0=b-d$ , or $b=d$ .

Substituting this in equations A, we get two distinct equations:

$ab=b+a$ and $-ab=b-a$ . Adding them results in $0=2b$ , or $b=0$ . With that $a=b=c=d=0$ .

Similarly we get the same result when substituting $b=-d$ in the original equations.

Subtracting Eqs (1) and (2), we have $(a-c)b = b+d-2c$ . Similarly, subtracting Eqs (3) and (4), we have $(a-c)d = (2a-b-d)$ . Adding these two eqs, we have $(a-c)(b+d)=2(a-c)$ . This implies that either $a=c$ or $b+d=2$ . Similarly manipulating other equations, we have either $b=d$ or $a+c=2$ .

If $a=c$ , then from Equ (1), we have $(a-1)b = a \implies b = 0 \implies c = 0 \implies a = 0$ . From Equ (2), for this case, we have $d=0$ . So if $a=c$ , then the only solution is $a=b=c=d=0$ .

If $b+d=2$ , then adding Equ (1) and Equ (4), we have $a(b+d)=a-c \implies 2a=a-c \implies a = -c$ . From Equ (1), letting $c=-a$ , we have $b = {a \over a-1}$ . Also from Equ (3), we have $d={a \over a+1}$ . Adding, we get ${a \over a-1} + {a \over a+1} = 2 \implies {2a^2 \over a^2-1} = 2$ which is impossible.

Similar, the case $a+c=2$ is also impossible.

So the only solution is $a=b=c=d=0$ .

We label the equations $(1),(2),(3),(4)$ in the order they are given for easier reference.

Adding all the equations together yields $ab+bc+cd+ad = 0 \to (a+c)(b+d) = 0 \to a = -c$ or $b = -d$ .

We let $a = -c$ . Then $(1)$ becomes $ab = b + (a)$ and $(2)$ becomes $b(-a) = (-a) - d \to ab = a + d$ . Thus, we have that $ab = a + b = a + d \to b = d$ , and so $(4)$ becomes $a(b) = a - b$ .

Subtracting $(4)$ ( $ab = a - b$ ) from $(1)$ ( $ab = a + b$ ) yields $ab - ab = b + a - (a - b) \to 2b = 0 \to b = 0$ , and so $b = d = 0$ . Finally, plugging $b = 0$ back in to $(1)$ gives $a\cdot(0) = (0) - c \to c = 0$ , and so $a = 0$ .

This tells us that the ordered quadruple $(0,0,0,0)$ is a solution to the given system of equations. If we now let $b = -d$ , we see that we can do essentially the same manipulations (but focusing on $(2)$ and $(3)$ rather than $(1)$ and $(4)$ ) to again obtain only $(0,0,0,0)$ as a possible solution to the system of equations.

Therefore, the only ordered quadruple of complex numbers $(a,b,c,d)$ that satisfies the given system of equations is $(0,0,0,0)$ , and so our answer is $\fbox{1}$ .

Considering the symmetry of the equations, we are inspired to add them, yielding $ab + bc + cd + ad = (a + c)(b + d) = 0.$ This means that at least one of $(a + c)$ or $(b + d)$ is 0. Based on the previous, we generate the following cases: $\text{case 1: }(a + c) = 0 \Longrightarrow a = - c\text{,}$ $\text{case 2: }(b + d) = 0 \Longrightarrow b = - d.$ Substituting $a = - c$ from case 1 into the first and second original equations, we obtain $ab = a + b \text{ and } ab = a + d.$ Subtracting the two new equations gives $(a + b) - (a + d) = (b - d) = ab - ab = 0$ , thus $b = d$ . We then add the first and second original equations; we have, $b(a + c) = b - d = 0.$ We note that if $b = 0$ , then $a = b = c = d = 0$ , and likewise when $a + c = 0$ . We solve for case 2 in a similar fashion and obtain the same result as in case 1. Therefore, the only solution of complex numbers $(a, b, c, d)$ that satisfy the system of equations is $(0, 0, 0, 0)$ , giving only $\fbox{1}$ ordered quadruple.