IMO

# IMO 2017 Q2

The problem: IMO 2017 Q2

Find all functions ${f: \mathbb R \rightarrow \mathbb R}$ such that for all ${x,\,y \in \mathbb R}$ we have the following:

$\displaystyle f(f(x)f(y)) + f(x+y) = f(xy).$

Solution and Discussion

Denote the given equation by ${P(x,y)}$. Cmon. It isn’t easy to see what solution would work save for ${f(x) \equiv 0}$. An observation is that if ${f(0) = 0}$ we simply substitute ${y=0}$ into the equation and we get ${f(x)=0}$ for all ${x.}$ So now we can assume ${f(0) \neq 0}$. Where to go?

It’s a medium so preliminary investigations can guide us quite a long way most likely. (Even for the monstrous 2015 Q5 one should at least get to showing ${f(0)=0}$ or ${2}$ before getting helplessly stuck.) So suppose ${x+y=xy}$, which is to say that ${y=\frac{x}{x-1}}$ , we shall have

$\displaystyle f(f(x)f(\frac{x}{x-1}))=0$

for all ${x \neq 1}$. Now suppose that for a certain ${\lambda \in \mathbb R}$ we have ${f(\lambda) = 0}$ (you see such ${\lambda}$ just has to exist, say ${\lambda = f(0)^2}$) we will then have ${f(f(\lambda)f(\frac{\lambda}{\lambda - 1})) =0}$ and thus ${f(0)=0}$ which is bad…. Unless ${\lambda = 1}$.

That is to say,

$\displaystyle f(\lambda) = 0 \Longleftrightarrow \lambda =1.\,\,\,\,\,(1)$

As a side, ${f(x)f(\frac{x}{x-1}) = 1}$ for all ${x \neq 1}$. Ergo, ${f(0)^2 = f(2)^2 =1}$. Also, ${P(1,1)}$ gives us ${f(0)+f(2)=0}$. Combining these two equations we have either

$\displaystyle f(0)=1 \,\mbox{ and }\,f(2)=-1,\,\,\,\,\,\,(*)$

or vice versa.

Do they really correspond to solutions at all? At present stage it seems that the equation (*) is suggesting that ${f(x) = 1-x}$, and the ‘vice versa’ demands ${f(x) = x-1}$. Substitute them in … they work! We want to only pursue one line of reasoning though.

Observation: if ${f(x)}$ is a solution, ${g(x) = -f(x)}$ is also a solution. This is because ${g(g(x)g(y))+g(x+y) = -(f(f(x)f(y))+f(x+y))=-f(xy)=g(xy)}$. So we can really WLOG assume (*) is true. (why not the vice versa? Because ${f(x)=1-x}$ is an involution, that is, ${f(f(x))=x,\,\forall x}$ which could simplify our work a lot potentially.)

First of all, ${P(x,1) \implies}$

$\displaystyle 1 + f(x+1) = f(x)\,\,\,\,\,\,\,(2)$

And here you go, ${P(x,0) \implies}$

$\displaystyle f(f(x))+f(x) = 1 \,\,\,\,\,\,\,(3)$

which is to say that ${f(x) = 1-x ,\, f(f(x))=x,\, \forall x \in f(\mathbb R).}$ So it remains to show ${f}$ is surjective!

If surjection doesn’t work, one might still try injection. So let’s try to show injection. Doesn’t seem easy… Suppose ${f(x_1)=f(x_2)}$, ${P(x_1, -x_2)}$ and ${P(-x_1,x_2)}$ gives us that, provided that ${f(-x_1) =f (-x_2)}$, ${f(x_1-x_2)=f(x_2-x_1)}$. So let’s pause for a moment, and see if ${f(-x_1)=f(-x_2)}$ can be shown…

As a matter of fact, your OP got exceedingly lucky when trying this problem. He somehow made a bad mistake that gave him true conclusions. Don’t ask him how. The OP just somehow figured out that

$\displaystyle f(-f(f(x)))=f(-x)\,\,\,\,\,\,\,\,(4)$

To show (4), we look back into the original equation ${P(x,y)}$.

$\displaystyle P(f(x),2) \implies \\ f(-f(f(x))) + f(f(x)+2) = f(2f(x))$

$\displaystyle P(x,-1)\implies \\ f(2f(x))+f(x-1)=f(-x)$

Combining the above two pieces of computation we get

$\displaystyle f(-f(f(x))) - f(-x) = -f(x-1) - f(f(x)+2) = -f(x) -1 - (f(f(x))-2) = 1- (f(f(x))+f(x)) = 0$

by referencing (2) and (3).

Whooosh. That was precarious. But now we have (4), which gives us not only that ${f(-x_1) = f(-x_2)}$, but also that, if ${k = x_1 - x_2}$, ${f(k) = f(-k) = f(-f(f(k)))}$, and if we ${f}$ both sides, ${f(f(k)) = f(f(-f(f(k))))}$. Suppose we could show that ${-f(f(k)) \in f(\mathbb R)}$, we would be able to ‘undress’ the RHS and get ${f(f(-f(f(k))))=-f(f(k))}$ and thus getting ${f(f(k))=0}$. Sounds like a plan. Well, (3) says if ${u \in f(\mathbb R)}$, ${1-u \in f(\mathbb R)}$ and (2) says if ${1-u \in f(\mathbb R)}$, ${-u \in f(\mathbb R)}$. Now ${f(f(k)) \in f(\mathbb R)}$ so it works.

Hooray! We have just shown that if ${f(k)=f(-k)}$ with ${k \neq 0}$, we must have ${f(f(k))=0}$ and thus ${f(k)=f(-k) = 1}$. Reference (2), we have that ${f(k+1)=0}$. Recall that ${1}$ is the only root to ${f(x)=0}$, we have ${k=0}$. Thus ${x_1=x_2}$. YAY!!! ${f}$ is indeed injective!!!

The rest is easy. (3) shows that ${f(f(f(x)))=f(x)}$ and by injectivity ${f(f(x))=x}$ and by (3) we have ${f(x) = 1-x}$ for all ${x \in \mathbb R}$.

The solutions are ${f(x)=x-1,\,f(x)=1-x,}$ and ${f(x)=0.}$

Comment

What a tedious problem…. Took me two hours + ${\varepsilon}$ to solve. It’s easy to leave gaps in the solution here and there, for instance the OP believed he got the identity that ${f(x+f(y))=f(y+f(x))}$ but in fact it demands ${x,y \in f(\mathbb R)}$ to hold. It helped him discovering (4) though. (4) is really important (and somewhat natural to think of after we have discovered (3)).

The entire solution shows the power of setting ${f(0)=1}$ instead of ${-1}$ because (3) will take some much more deplorable forms had we chosen ${f(0)=-1}$ to work with, so will be (4) — in my humble opinion. I am happy to be shown wrong and be presented a proof using ${f(0)=-1}$ as starting point — I have not yet tried, but it looks disastrous to me.

OP was surprised to see that the key to the solution is injectivity really. The equation looks nothing like something for which injectivity would be a step towards solution — no naked variable anywhere. The point that ${1}$ is the only root to ${f(x)=0}$ must be made for it is the essential cornerstone of any effort towards injectivity.

One might be tempted to use analysis, but I personally do not believe that analysis will be helpful. Analysis required inequality whereas this problem only gives equality that is difficult to be transformed into inequalities.