On IMO2017 P2

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I did IMO 2017 P2 with my friends yesterday. Now I would like to share my thoughts on this problem.

Problem: Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that, for any real numbers $x$ and $y$ ,

$f(f(x)f(y))+f(x+y)=f(xy)$

The initial step must be taking certain values, testing its properties and then start guessing the solution. Well, obviously, the constant function $f(x)=0$ is a solution to the problem.

Now we can try to take $x=y=0$ , giving us

(1) $\begin{equation*} f(f(0)\cdot f(0)))+f(0)=f(0)\end{equation*}$

and

(2) $\begin{equation*} f(f(0)f(0)))=0\end{equation*}$

Now if we take $x=x, y=0$ , we could get

(3) $\begin{equation*} f(f(x)f(0)))+f(x)=f(0)\end{equation*}$

$f(0)$ appears on both side, and it is quite hard to analyze it. What if we already know the value of $f(0)$ ? Suppose that $f(0)=0$ , then $f(0)+f(x)=f(0)$ , so $f(x)=0$ , giving us the trivial solution. Now suppose that $f(0)\neq 0$ , notice that if we replace $f$ with $-f$ , then $-f$ is also a solution. That means we can only consider the case where $f(0)>0$ .

Now we are stuck. Maybe we should try to get more informations. One strategy we often use involving solving functional equations is cancelation. Let $x+y=xy$ , we can replace $y=\dfrac{x}{x-1}$ . Plugging in, we get

(4) $\begin{equation*} f(f(x)f(\frac{x}{x-1}))=0\end{equation*}$

This result could be pretty useful, as we can use it to analyze the zeros of $f(x)$ . We know that there must have at least one zero. Now suppose that for some $f(a)$ , $f(a)=0$ , then $f\left(f(a)f(\frac{a}{a-1})\right)=f(0)>0$ , which is a contradiction. That means if we have a zero, then it must be $1$ . Therefore, $f(1)=0$ , and $f(x)f(\frac{x}{x-1})=1$ whenever $n\neq 1$ .

Now we could more or less take a guess. It is impossible that $f(x)$ has a degree greater than one, and it is very likely that it is a linear function, which is $f(x)=1-x$ . We proceed with this conjecture.

Having $f(1)=0$ , we can plug $x=1$ into (4). We derive

(5) $\begin{equation*} f(0)^2 = 1\end{equation*}$

This gives us $f(0)=1$ . So we could try more values. Taking $x=x, y=1$ , we get

(6) $\begin{equation*} f(f(x)f(1))+f(x+1)=f(x)\end{equation*}$

(7) $\begin{equation*} f(x+1)=f(x)-1\end{equation*}$

We can also take $x=x, y=0$ , we get

(8) $\begin{equation*} f(f(x))+f(x)=f(0)=1\end{equation*}$

(9) $\begin{equation*} f(f(x))=1-f(x)\end{equation*}$

Viewing this composite function, we can use the traditional method.

(10) $\begin{equation*} f(1-f(x))=f(f(f(x))=1-f(f(x))=f(x)\end{equation*}$

The solution could be very similar to our conjecture. If $1-f(x)=x$ , then $f(x)=1-x$ . That means if we can prove injection, then we are done!

This could be a hard part. We go back to our problem

(11) $\begin{equation*} f(f(x)f(y))=f(xy)-f(x+y)\end{equation*}$

Suppose that $f(a)=f(b)$ , we need to pick specific $x$ and $y$ to give some useful information. Seeking relation to zero, we want $f(xy)-f(x+y)=1$ . So let $xy=a+1$ and $x+y=b$ . This is achievable, since by our previous result, $f(x+1)=f(x)-1$ . If we want the determinant $b^2-4(a+1)\geq 0$ , we can shift $a$ and $b$ so that $b$ is large enough. Finally, we get