2019 IMO Problems/Problem 1
Let be the set of integers. Determine all functions such that, for all integers and ,
Let us substitute in for to get
Now, since the domain and range of are the same, we can let and equal some constant to get Therefore, we have found that all solutions must be of the form
Plugging back into the original equation, we have: which is true. Therefore, we know that satisfies the above for any integral constant c, and that this family of equations is unique.
(This solution does not work though because we don't know that is surjective)
We plug in and to get respectively.
Setting them equal to each other, we have the equation and moving "like terms" to one side of the equation yields Seeing that this is a difference of outputs of we can relate this to slope by dividing by on both sides. This gives us which means that is linear. (Functional equations don't work like that unfortunately)
Let Plugging our expression into our original equation yields and letting be constant, this can only be true if If then which implies However, the output is then not all integers, so this doesn't work. If we have Plugging this in works, so the answer is for some integer
Solution 3: The only one that actually works
The only solutions are For some integer
Obviously these work. We prove these are the only linear solutions. Plug and separately to get that Plug to see and subtracting from both sides shows to be additive thus linear by Cauchy since this is on integers. Thus, is linear, and so we are done since it can be easily shown that and are the only linear solutions by plugging into the equation.
~ LLL2019 (first to post correct solution on here)
Solution 4: This works as well
We claim the only solutions are and for some integer ., which obviously work. Plugging in and give , so . Since this difference is constant and , we must have is linear (finite differences or induction). It is easy to see the only linear solutions are those specified above.
~ Ezra Guerrero