Trying to solve $\frac{f(x) f(y) - f(xy)}{3} = x + y + 2$ for $f(x)$

Question

Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that $$\frac{f(x) f(y) - f(xy)}{3} = x + y + 2$$ for all $x,y \in \mathbb{R}$. Find $f(x)$.

I started by multiplying both sides by $3$, which gets $$f(x)f(y)-f(xy)=3x+3y+6.$$

I tried to find something by substituting $y=0$, so $$f(x)f(0)-f(0)=3x+6.$$

However, I don't see anything useful. How would I continue on this problem.

Answer 1

Setting $x=y=0$ in the functional equation yields $$f(0)^2-f(0)-6=0 \implies f(0)=3 \text{ or } f(0)=-2$$ Set $y=0$ to obtain $$f(0)(f(x)-1)=3x+6$$ Now $f$ can be determined by substituting the values obtained for $f(0)$. Substituting back into the original functional equation tells us that only one of them is valid.

Finally, we have a unique solution for $f$: $$f(x) = x+3$$

Answer 2

Suppose $f$ is a solution. Do the susbsitution $y=0$ and now suppose $f(0) = 0$. This means that for all $x$, $0 = 3x+6$. This is absurd. Thus, you can conclude that if $f$ is a solution, $f(0) \neq 0$. So you can divide by $f(0)$ and find out that $f(x) = \frac{3x+6}{f(0)} +1$. Conversely, is this a solution? Compute for what value of the constant $f(0)$ it satisfies the desired relation.