Maybe it's too easy for this TST... Anyway, this is my solution: Let $ P(m,n) $ be the assertion of $ f(f(m)+f(n))=f(f(m))+f(n) $ $ P(m,1) \Rightarrow f(f(m)+2)=f(f(m))+2 \rightarrow f(2k)=2k+2 \forall k \in \mathbb{N} $ ((Just a induction~ Notice we have $ f(c) $ is odd $ \Leftrightarrow c $ is odd since $ f $ is a injective function. Now, let $ a $ be the smallest odd number that $ f $ could take and $ b $ s.t $ f(b)=a $ Then we can get $ f(2k+1)=2k+1 $ $ \forall 2k+1 \ge a \rightarrow b<a $ Therefore, $ \{f(b),f(a),f(a+2),...\} $ covers all odd value taken by $ f $ which implies $ b=3, a=5 \Rightarrow f(n)=n+2 $ $ \forall n \ge 2 $!!