If $f(3x)=f(x)+f(3)$, prove that

Question

If $$f(3x)=f(x)+f(3),$$prove that : $$f(1)=0\\f(3)=0\\f(9)=0\\f(27)=0$$

My attempt: Here: $$f(3x)=f(x)+f(3)$$ If $$x=1$$

$$f(3\times 1)=f(1)+f(3)$$ $$f(1)=0$$. I got the first one but how should I prove the rest?

Answer 1

$$f(3x)=f(x)+f(3)$$

For $x=0$ we get $f(0)=f(0)+f(3) \Rightarrow f(3)=0$.

For $x=1$ we get $f(3)=f(1)+f(3) \Rightarrow f(1)=0$.

For $x=3$ we get $f(9)=2f(3)=0$.

For $x=9$ we get $f(27)=f(9)+f(3)=3f(3)=0$.

Answer 2

$$f(0)=f(3\times 0)=f(0)+f(3)\to f(3)=0$$$$f(9)=f(3\times 3)=f(3)+f(3)=0,f(27)=f(3\times 9)=f(9)+f(3)=0$$

Answer 3

First show that, taking $x= 0$, $f(3 \times 0)= f(0)= f(0)+ f(3)$ so $f(3)= ?$.
Now, $f(9)= f(3\times 3)= f(3)+ f(3)$ and $f(27)= f(3\times 9)= f(9)+ f(3)$.