Finding all $ f : \mathbb R \to \mathbb R $ satisfying $ f \bigl ( f ( x ) f ( y ) \bigr ) + f ( x + y ) = f ( x y ) $ for all $ x , y \in \mathbb R $

Solution 1:

$$ f(f(x)f(y))+f(x+y)=f(xy). \ \ \ (1)$$ Let $f(0)=c$. Choose $x=y=0$, we can get $$f(c^2)=0.\ \ \ (2)$$ Choose $y=0$, we can get $$f(cf(x))+f(x)=c.\ \ \ (3)$$ Choose $y=\frac{x}{x-1}(x\neq 1)$, we can get
$$f\left(f(x)f\left(\frac{x}{x-1}\right)\right)=0(x\neq 1).\ \ \ (4)$$

When $c=0$, from equation $(3)$ we get $f(x)=0.$

When $c\neq 0$, i.e. $f(0)\neq 0$. Form equation $(4)$ we know there esists $x_0\neq0$ such that $f(x_0)=0.$

We claim that :$x_0=1$.

Otherwise, choose $x=x_0$ in equation $(4)$, we get $f(0)=0$ which is a contradiction.

Combining equation $(2)$, we know $c=1$ or $-1$.

If $c=1$, that is to say $f(0)=1,$ choose $y=1$ in equqtion $(1)$, we get $f(x+1)=f(x)-1$. So $f(n)=1-n$ and $f(x+n)=f(x)-n$ for all $n\in Z$. By equation $(1)$ we get $$f(f(x)f(y)+1)+f(x+y+n)=f(xy+n+1).\ \ \ (5)$$ In the following we prove that: $f$ is injective. If $f(a)=f(b)$, choose integer $n$ such that $(b-n)^2>4(a-n-1)$, such that there exists $x_0,y_0$ statisfying $$x_0y_0+n+1=a,x_0+y_0+n=b.$$ From equation$(5)$ we get $f(x_0)f(y_0)+1=1$, so $f(x_0)=0$ or $f(y_0)=0$ ,i.e. $x_0=1$ or $y_0=1$, and this implies $a=b$ which is the injectivity of function $f$.

From equation $(3)$, we know $f(f(x))=1-f(x)$. On the one hand, $f(f(f(x)))=1-f(f(x))=1-(1-f(x))=f(x);$ on the other hand, $f(f(f(x)))=f(1-f(x))$. Injectivity of $f$ implies $f(x)=1-x.$

If $c=-1$, we can get $f(x)=x-1$ in the same way as above!

In conclusion, all the solutions of the fucntioanl equation are the following: $$f(x)=0; f(x)=1-x; f(x)=x-1.$$