
6. Let $\mathbf{R}$ denote the set of all real numbers. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfying the condition

$$
f(x+y)=f(x) f(y) f(x y)
$$

for all $x, y$ in $\mathbf{R}$.


Solution: Putting $x=0, y=0$, we get $f(0)=f(0)^{3}$ so that $f(0)=0,1$ or -1 . If $f(0)=0$, then taking $y=0$ in the given equation, we obtain $f(x)=f(x) f(0)^{2}=0$ for all $x$.

Suppose $f(0)=1$. Taking $y=-x$, we obtain

$$
1=f(0)=f(x-x)=f(x) f(-x) f\left(-x^{2}\right)
$$

This shows that $f(x) \neq 0$ for any $x \in \mathbf{R}$. Taking $x=1, y=x-1$, we obtain

$$
f(x)=f(1) f(x-1)^{2}=f(1)[f(x) f(-x) f(-x)]^{2}
$$

Using $f(x) \neq 0$, we conclude that $1=k f(x)(f(-x))^{2}$, where $k=$

$f(1)(f(-1))^{2}$. Changing $x$ to $-x$ here, we also infer that $1=k f(-x)(f(x))^{2}$. Comparing these expressions we see that $f(-x)=f(x)$. It follows that $1=k f(x)^{3}$. Thus $f(x)$ is constant for all $x$. Since $f(0)=1$, we conclude that $f(x)=1$ for all real $x$.

If $f(0)=-1$, a similar analysis shows that $f(x)=-1$ for all $x \in \mathbf{R}$. We can verify that each of these functions satisfies the given functional equation. Thus there are three solutions, all of them being constant functions.

