Problem : (ISL 2009)
Find all functions from the set of real numbers into the set of real numbers which satisfy for all , the identity
Proposed by Japan

Solution :

Let be the assertion

Lemma 1 :
Proof :
Let
Consider
That is
But this satisfy the conditions of the given equation.
Hence

Lemma 2 :
Proof :
Let if possible their exist some such that
Consider
A contradiction !!!

Thus we must have only root of as .


Lemma 3 : is injective
Proof :
Let if possible there exist some and
Then consider
And now consider

By Lemma 2,
Thus we have either or which again implies (by Lemma 2)
A contradiction.
Thus we get is injective.

Lemma 4 :
Proof :
Just consider

Lemma 5 :
Proof :
For this is obvious.
Now consider some
By Lemma 4,
Using injectivity,
Hence the lemma.

Lemma 6 :
Proof :
Consider
Injectivity implies
Replacing by ,
Putting this back,

Lemma 7 :
Proof :
Let be the assertion

Consider
We have,
That is, we have
Setting here, we have

Hence we have

Lemma 8 :
Proof :
Consider
Now consider

Now using is odd (Lemma 5), and adding the two equations,
That is

Main Proof :

Since , we will form cases
Also observe that by lemma 6, ,

Case 1 :

Since is odd,
Also,
Lemma 8 yields,
That is,

Consider and using ,
Usinf injectivity

That is
This is a solution.

Case 2 :

This yields and
Lemma 8 yields
That is

Again consider and use
Using injectivity,


That is



So the only solutions are and