Solving a functional equation over $\mathbb{R}^+$

Question

How do we solve the functional equation $f : \mathbb{R}^{+}\rightarrow \mathbb{R}^+$ such that $f(x)f(y)+f(xy) = xy$?

Answer 1

Notice that for $y=1$, one gets:

$$f(x)f(1)-f(x) = \left(x+2\right)\left(\frac{1}{x}+1\right) \Rightarrow f(x) = \frac{1}{f(1)-1}\left[\left(x+2\right)\left(\frac{1}{x}+1\right)\right].$$

Now, observe that:

$$f(1) = \frac{1}{f(1)-1}\left[\left(1+2\right)\left(\frac{1}{1}+1\right)\right] = \frac{6}{f(1)-1}.$$

Then:

$$f(1)(f(1)-1) = 6 \Rightarrow f^2(1) - f(1)-6 = 0 \Rightarrow f(1)= 3 \vee f(1) = -2.$$

Since $f:\mathbb{R}^+ \to \mathbb{R}^+$, then $f(1) = 3.$

Hence:

$$f(x) = \frac{1}{2}\left(x+2\right)\left(\frac{1}{x}+1\right).$$

Answer 2

For $x=y=1$ you get $f(1)^2-f(1)=6\iff f(1)=3$, since the other possible solution is negative (namely, $-2$).

Now for any $x>0$, taking $y=1$ one gets $$f(x)f(1)-f(x)=(x+2)(1/x+1)\iff f(x)=\frac{(x+2)(1/x+1)}{f(1)-1}=\frac{(x+1)(x+2)}{2x}$$

Answer 3

If $y=1$ then $f(x)f(1)+f(x)=x$

$f(x)(f(1)+1)=x$

if $f(1)+1=0$ then $0=x$ for all x which is not true so $f(1)+1\neq0$

$f(x)=\frac{x}{f(1)+1}=\frac{x}{c}$ where c is a constant for all x.

$(\frac{x}{c})(\frac{y}{c})+\frac{xy}{c}=\frac{xy}{c^2}+\frac{xy}{c}=xy$

$\frac{1}{c^2}+\frac{1}{c}=1$

$1+c=c^2$

c has to be positive as it is positive to positive so this is the well known golden ratio $\varphi$

$f(x)=\frac{x}{\varphi}$

Substitute to confirm solution.