IMO 2019 in South Africa

 Let Z be the set of integers. Determine all function `f: Z \rightarrow Z` such that, for all integers `a` and `b` 

                        `f(2a)+2f(b)=f(f(a+b))`                      `(1)`

Answer: The solution are `f(n)=0` and `f(n)=2n+k` for any constant `k\in Z`

Substituting `a =0, b= n+1 ` gives  `f(f(n+1))=f(0)+2f(n+1)`. 

Substituting `a =1, b= n` gives `f(f(n+1))=f(2)+2f(n)` . 

    In particular, `f(0)+2f(n+1)=f(2)+2f(n)`, and so  `f(n+1)-f(n)=1/2(f(2)-f(0))` .

Thus `f(n+1)-f(n)` must be constant. Since `f` is defined only on `Z`, this tells us that f must be a linear function; write`f(n)= Mn+ K` for arbitrary constants `M` and `K`, and we need only determine which choices of `M` and `K` work.

 Now, (1) becomes 

                        `2Ma +K + 2(Mb + K)= M(M(a+b)+ K)+K` 

which we may rearrange to form

                         `(M-2)( M(a+ b)+K)=0`. 

    Thus, either `M= 2`, or `M(a+b)+ K= 0` for all values of `a+b`. In particular, the only possible solutions are `f(n)=0` and `f(n)=2n +K` for any constant `K \in Z`, and these are easily seen to work.