Find all functions such that, for all integers that satisfy , the following equality holds:

(Here denotes the set of integers.)

Solution:

So I thought this problem was pretty fun... but I've liked functional equations ever since I got lucky on USAMO 2014 #2 so I'm biased. Even so, it was so tedious that I kind of gave up at the end

Lemma 1:

Proof: Letting yields that which implies the desired result.

Lemma 2: for all

Proof: Letting and applying Lemma 1 yields that and by the equality case in AM-GM we have the desired result.

Now, letting and applying Lemma 2 we find that for all Particularly, either or .

Case 1:

Then by letting for and using a quick induction we find that for all . Now, consider any two odd integers and . Letting we have that which by the equality case in AM-GM implies that . Therefore we have found a solution

Case 2:

Now letting we find that .

Case 2.1:

Letting we find that .

Case 2.1.1:

Now by a similar induction to the one used in Case 1 we have that for all . Now consider any two integers and that satisfy . Letting yields which by the equality case in AM-GM implies that Now, consider any two odd numbers and . Letting we have that which by the equality case in AM-GM implies that By induction the values of evaluated at odd integers are equal. By letting to obtain the final restriction that , we have found a solution

Case 2.1.2:

Then . Letting we find that so and we default to Case 1 (in this case, we actually have ).

Case 2.2:

I will show that for all by strong induction on Letting and repeatedly applying the induction hypothesis we have that If for some then letting we find that . If then we default to Case 1. If then Letting we find that . If then letting we find by the AM-GM stuff that and this becomes similar to past cases. Similarly if we proceed similarly to Case 2.1.2. Therefore the only new solution is

These three solutions can all be checked to work so we are done.