Find all functions $f\colon \mathbb R\rightarrow \mathbb R$ such that for all reals $x$ and $y$: $$f(f(x)f(y))+f(x+y)=f(xy).$$

It was six hours ago in IMO 2017 (problem 2).

I tried the standard way: $x=0$, $x=y$, $x=1$,etc. but without any success.

Let $f(0)=a$. Hence, for all $x\in\mathbb R$ we have $f(a(f(x))+f(x)=a$. What is the rest?

Also we have $(x-1)(y-1)-1+x+y-1=xy-1$ and



$$ f(f(x)f(y))+f(x+y)=f(xy). \ \ \ (1)$$ 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: $$f(x)=0; f(x)=1-x; f(x)=x-1.$$

  • 3
    $\begingroup$ It is a beautiful problem (, but f*** it). Somehow this is the only solution online. One of the syrian guys came up with a soultion using substitution. $\endgroup$ Aug 9 '17 at 22:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.