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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

$1-(1-x)(1-y)+1-x-y=1-xy$.

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  • $\begingroup$ what do you have so far for $f(0), f(1)?$ $\endgroup$ – Doug M Jul 18 '17 at 18:11
  • $\begingroup$ $x\mapsto 0$ is a solution. $\endgroup$ – hamam_Abdallah Jul 18 '17 at 18:16
  • $\begingroup$ The rest is $f (a^2)=0$. $\endgroup$ – hamam_Abdallah Jul 18 '17 at 18:18
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    $\begingroup$ $f(f(0)^2)+f(0)=f(0)\implies f(f(0)^2)=0.$ $\endgroup$ – mfl Jul 18 '17 at 18:24
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    $\begingroup$ @астон вілла олоф мэллбэр See here: artofproblemsolving.com/community/c6h1480146 $\endgroup$ – Michael Rozenberg Jul 18 '17 at 18:40
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$$ f(f(x)f(y))+f(x+y)=f(xy). \ \ \ (1)$$ 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: $$f(x)=0; f(x)=1-x; f(x)=x-1.$$

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    $\begingroup$ 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. $\endgroup$ – Frieder Jäckel Aug 9 '17 at 22:12

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