2019 IMO Problems/Problem 1
Revision as of 16:46, 18 July 2019 by Euclid geometry (talk | contribs) (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.)
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.
