2016 APMO Problems/Problem 5
Find all functions such that for all positive real numbers .
We claim that is the only solution. It is easy to check that it works. Now, we will break things down in several claims. Let be the assertion to the Functional Equation.
Claim 1: is injective.
Proof: Assume for some . Now, from and we have:
Now comparing, we have as desired.
This gives us the power to compute . From we get and injectivity gives . Showing that is unbounded above is also easy as we can fix and let blow up to in the original Functional equation..
Claim 2: is surjective.
This gives that . Putting , we get . By induction, surjectivity is proved as and we are essentially done.
Now, we have claim that if for some , then . This can be achieved by putting , , and . Let us calculate
Everything is fine, but we would have show that such numbers do exist. We will show this in the next lemma.
Lemma 1: There exits such that Proof: Notice that is surjective. So, we will solve for in the system of linear equation. For convinience, let for as is surjective. Solving we get: Now, you can chose such that and is positive. This finishes the proof to the lemma.
Claim 3: obeys Jensen's Functional Equation,i.e., Proof: