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?

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:
$$(a,b,c,d)=(-1, -1, -1, -1), (-1, 1, -1, 1), (-1, 1, 1, -1), (-1, -1, 
    1, 1), (1, 1, -1, -1), (1, -1, -1, 1) , (1, -1, 1, -1), (1, 1, 1, 1).$$
