# 2008 IMO Problems/Problem 4

## Problem

Find all functions (so is a function from the positive real numbers) such that

for all positive real numbers satisfying

## Solution

Considering and which satisfy the constraint we get the following equation:

At once considering we get and knowing that the only possible solution is since is impossible.

So we get the quadratic equation:

Solving for as a function of we get:

At once we see that for one value of , can only take one of 2 possible values:

.

Take into consideration that but verifies the quadratic equation and thus so far we can't say that or alternatively . This is indeed the case but we haven't proved it yet.

To prove the previous assertion consider 2 values such that while having

Consider now the original functional equation with which verifies the constraint. Substituting we have:

Now either or . (notice that by hypothesis)

If then we have and since the only solution is .

If then we have and since the only solution is .

So the only solutions are or in which case both alternatives imply . Thus we conclude that solutions to the functional equation are a subset of .

Finally, plug each of these 2 functions into the functional equation and verify that they indeed are solutions.

This is trivial since is an obvious solution and for we have:

provided that which verifies the original constraint.

So the functional equation has 2 solutions:

## Video Solution

https://youtu.be/wb2gp8uoGfM [Video Solution by little fermat]