How many ordered triples $(a, b, c)$ of non-zero real numbers have the property that each number is the product of the other two?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$
We have $ab = c$, $bc = a$, and $ca = b$, so multiplying these three equations together gives $a^{2}b^{2}c^{2} = abc \implies abc(abc-1)=0$, and as $a$, $b$, and $c$ are all non-zero, we cannot have $abc = 0$, so we must have $abc = 1$. Now substituting $bc = a$ gives $a(bc) = 1 \implies a^2 = 1 \implies a = \pm 1$. If $a = 1$, then the system becomes $b = c, bc = 1, c = b$, so either $b = c = 1$ or $b = c = -1$, giving $2$ solutions. If $a = -1$, the system becomes $-b = c, bc = -1, -c = b$, so $-b = c = 1$ or $b = -c = 1$, giving another $2$ solutions. Thus the total number of solutions is $2 + 2 = 4$, which is answer $\boxed{\text{D}}$.