Number of distinct values Question:
How many possible values of (a, b, c, d), with a, b, c, d real, are there
such that abc = d, bcd = a, cda = b and dab = c?
I tried multiplying all the four equations to get:
$$(abcd)^2 = 1$$
Not sure how to proceed on from here. Won't there be infinite values satisfying this equation?
 A: If one of the variables is $0$ then all of them are...so one solution is $\{0,0,0,0\}$. 
Let's now exclude that solution.  As the OP points out, multiplying the equations together yields $$(abcd)^3=abcd\implies (abcd)^2=1$$
Now, suppose that $a\neq \pm 1$.  We remark that $$abcd=a\times bcd=a\times a=a^2\neq \pm 1$$ which contradicts $(abcd)^2=1$.  Thus $a$ must be $\pm 1$.  Similarly, each variable must be $\pm 1$.  Inspection quickly shows that either all four have the same sign or we have two of each, and we are done.
A: 1) If $abcd = 0$, i.e., if one at least among $a,b,c,d$ is zero, all are clearly zero. Thus there is a solution:
$$a=b=c=d=0$$
2) If $abcd \neq 0$, let $A=\ln(|a|), B=\ln(|b|), C=\ln(|c]), D=\ln(|d|).$
Taking absolute values and then logarithms of the 4 equations, we obtain the following linear  homogeneous system
$$\begin{bmatrix}1&1&1&-1\\-1&1&1&1\\1&-1&1&1\\1&1&-1&1 \end{bmatrix}\begin{bmatrix}A\\B\\C\\D \end{bmatrix}=\begin{bmatrix}0\\0\\0\\0 \end{bmatrix}$$
The determinant of the matrix $M$ of the system is $16 \neq 0$, thus the kernel of $M$ is reduced to $$(A,B,C,D)=(0,0,0,0)=(\ln(1),\ln(1),\ln(1),\ln(1))$$
Therefore 
$$|a|=1, |b|=1, |c|=1, |d|=1 \ \ \ (1)$$
All solutions of (1) may not be solutions to the initial system (because, by taking absolute values, we possibly have enlarged the set of solutions).
Thus, we have to check the 16 different possible sign combinations for $a,b,c$ and $d$. Doing this, 8 solutions remain:
$$(a,b,c,d)=(-1, -1, -1, -1), (-1, 1, -1, 1), (-1, 1, 1, -1), (-1, -1, 
    1, 1), (1, 1, -1, -1), (1, -1, -1, 1) , (1, -1, 1, -1), (1, 1, 1, 1).$$
A: $$(abcd)^3=abcd \\ \implies (abcd)^2=1\ or, \ abcd=0 \\ \implies  abcd=+1,-1,0 $$
If we look carefully, the initial conditions clearly say that we can't take $3$,$-1$s or only one $-1$. from these, one solution is $(a,b,c,d)=(0,0,0,0)$ and for the other ones i.e. for  $abcd=1$  we can say , no of pairs are $\frac{4!}{2! \times 2!}=6 \ (2--(+1)\ and \ 2--(-1))$, $(a,b,c,d)=(1,1,1,1),(-1,-1,-1,-1)$. Hence, total $9$ solutions. $\blacksquare$
