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))` .
