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 .