2005 IMO Shortlist Problems/A2
Contents
[hide]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
,
![$[f(x)]^2 = f(x)f(x) = 2f(z) \ge 2 \cdot 2^{1-\frac{1}{2^n}} = 2^{2\left(1 - \frac{1}{2^{n+1}} \right) }$](http://latex.artofproblemsolving.com/9/6/1/961756ada48786d9880fae1512167c09f2d52a3b.png)
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.