Let Z be the set of integers. Determine all function f:Z→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∈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)=12(f(2)-f(0)) .
