Difference between revisions of "2019 IMO Problems/Problem 1"

(This is Problem 1 of the 2019 IMO along with one solution. Please feel free to add more solutions which may be more elegant, and to correct this one if it is deemed incorrect.)
(No difference)

Revision as of 16:46, 18 July 2019

Problem:

Let Z be the set of integers. Determine all functions f : Z → Z such that, for all integers a and b, f(2a) + 2f(b) = f(f(a + b))

Solution 1:

Let us substitute 0 in for a to get: f(0) + 2f(b) = f(f(b))

Now, let x = f(b) to get and f(0) equal some constant c: c + 2x = f(x). Therefore, we have found that all solutions must be of the form f(x) = 2x + c.

Plugging back into the original equation, we have: 4a + c + 4b + 2c = 4a + 4b + 2c + c which is true. Therefore, we know that f(x) = 2x + c satisfies the above for any integral constant c, and that this family of equations is unique.