# 2005 IMO Shortlist Problems/A2

## Problem

(*Bulgaria*)
Let denote the set of all postive real numbers. Determine all functions such that

for all positive real numbers and .

## Solutions

### Solution 1

**Lemma 1.** is non-decreasing.

*Proof.* Suppose, on the contrary, that there exist such that . We set , so , to obtain

,

or , a contradiction. ∎

**Lemma 2.** is not strictly increasing.

*Proof.* Assume the contrary. Then for all we have

so . Furthermore, since is injective, we have , which implies , or

But then for arbitrarily close to 0, becomes less than 2, a contradiction. Thus is not strictly increasing. ∎

Now, let , . If is in the interval , then , so

so . It follows that

,

so , for all in that interval.

But if , then setting in the original equation gives , by induction. It follows that for any , there exist and such that , and since is non-decreasing, we must have for all . It is easy to see that this satisfies the given equation. Q.E.D.

### Solution 2

**Lemma 1.** There exist distinct positive such that .

*Proof.* Suppose the contrary, i.e., suppose is injective. Then for any , we have

,

which implies

.

This means that either and for all (a contradiction, since that is not injective), or , for some real . But setting then gives us a quadratic in with a nonzero leading coefficient, which has at most two real roots, implying that can only assume two different values, a contradiction. ∎

**Lemma 2.** There exist and infinitely many such that , for all nonnegative integers .

*Proof.* Let be the distinct positive reals of Lemma 1 such that ; without loss of generality, let . Letting yields

Since , this implies . We now prove that for all nonnegative , by induction. We have just proven our base case. Now, assume . Setting gives us

,

so , as desired. ∎

**Lemma 3.** For all , .

*Proof.* We note that is the solution to the equation , which is positive when , so if this is the case, setting to this value gives us

,

and since , this implies, , a contradiction. ∎

**Lemma 4.** .

*Proof.* We will prove by induction that , for all . Our base case comes from Lemma 3. Now, if for all , , then for some ,

so , as desired.

Now, suppose there exists some . Then there exists some such that . But this gives us , a contradiction. Thus for all , . ∎

We will now prove that is the only solution to the functional equation. Consider any value of . By Lemma 2, there exists some such that . Setting then gives us

But from Lemma 4, we know , so we must have . This constant function clearly satisfies the given equation. Q.E.D.

*Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.*