One more functional equation

Let $P(x,y)$ be the statement $f\left( {x - y} \right) = f\left( x \right) \cdot f\left( y \right) \cdot f\left( {\frac{x}{y}} \right)$ .

$P\left( {0,\;1} \right) \Rightarrow f\left( { - 1} \right) = {f^2}\left( 0 \right)f\left( 1 \right)\ \ \ \ \ \ \left( 1 \right)$

$P\left( {0,\; - 1} \right) \Rightarrow f\left( 1 \right) = {f^2}\left( 0 \right)f\left( { - 1} \right)\mathop \Rightarrow \limits^{\left( 1 \right)} f\left( 1 \right) = {f^4}\left( 0 \right)f\left( 1 \right) \Rightarrow \begin{cases}f(1)=0\\\text{or}\\f(0)=\pm 1\end{cases}$

If $f\left( 1 \right)=0$ , then $P\left( {x + 1,\;1} \right) \Rightarrow f\left( x \right) = {f^2}\left( {x + 1} \right)f\left( 1 \right) \Rightarrow f\left( x \right) = 0,{\text{ }}\forall x \in \mathbb{R}$ . Hence, $\boxed{f \equiv 0}$ is one solution, since it satisfies $P(x,y)$ .

If $f\left( 1 \right)\ne0$ , then $f\left( 0 \right)=\pm1$ . Exploring this case we have:

$$

$P\left( {1,\;1} \right) \Rightarrow f\left( 0 \right) = {f^3}\left( 1 \right)$ , thus, $f\left( 1 \right) = f\left( 0 \right) = 1$ , or $f\left( 1 \right) = f\left( 0 \right) = -1$ .

$$

In any case, for $x \ne 0$ , $P\left( {x,\;x} \right) \Rightarrow f\left( 0 \right) = {f^2}\left( x \right)f\left( 1 \right) \Rightarrow {f^2}\left( x \right) = 1,$ which is also true for $x=0$ .

$P\left( {x + 1,\;1} \right) \Rightarrow f\left( x \right) = {f^2}\left( {x + 1} \right)f\left( 1 \right) \Rightarrow f\left( x \right) = f\left( 1 \right)$ , for all $x \in \mathbb{R}$ .

The last leads us to two more solutions: $\boxed{f \equiv 1}$ , or $\boxed{f \equiv -1}$ , which both satisfy $P(x,y)$ .

To conclude, the answer is $\boxed{3}$ .

Putting $x=0$ we see that $f(-y) = f(0)^2f(y)$ for nonzero $y$ , which means that $f(0)^4f(y) = f(y)$ for all nonzero $y$ . Thus either $f(0)^4 = 1$ , so that $f(0) = \pm1$ , or else $f(y) = 0$ for all nonzero $y$ . But then $f(0) = f(1)f(1)f(1) = 0$ , so that $f \equiv 0$ (one solution)

Suppose that $f(0) = 1$ . Then $f(-y) = f(0)^2f(y) = f(y)$ for all nonzero $y$ , and hence for all $y$ . Also, putting $x=y$ we see that $1 = f(0) = f(x)^2f(1)$ for all nonzero $x$ . In particular $f(1)^3 = 1$ , so that $f(1) = 1$ , and hence $f(x)^2 = 1$ for all $x$ . But then $f(x) = f(\tfrac12x)f(-\tfrac12x)f(-1) = f(\tfrac12x)^2f(-1) = 1$ for all nonzero $x$ , and hence $f \equiv 1$ (a second solution).

If $f(0)=-1$ $g(x) = -f(x)$ is another solution of the functional equation with $g(0) = 1$ , so that $g \equiv 1$ as above. Thus $f \equiv -1$ (the third and final solution).

Thus here are exactly $\boxed{3}$ solutions, all constant.

In the case of constant functions $f(x) = C$ , we have $C = C^3 \Rightarrow C^3 - C = C(C+1)(C-1) = 0 \Rightarrow C = 0, \pm1$ .