### IMO 2011, Problem 3

IMO Problem 3: Let a function satisfy the relation . Prove that for , .

Let . Then we have . Cancelling on both sides, we get .

For , if , then the above gives a contradiction, as but .

Now we see that . This is because from the above equation, we have . This is possible only if .

Hence so far, we have that for , . Now we shall prove that for all , we have . This will give us the desired contradiction.

We have . Let . Then we have , or . This, coupled with the fact that for we have , we get that for all , we have .