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?

$$(abcd)^3=abcd \\ \implies (abcd)^2=1\ or, \ abcd=0 \\ \implies  abcd=+1,-1,0 $$
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. $\blacksquare$
