Why Is This Function Bijective?

Question

Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that $$f(f(f(n))) + f(f(n)) +f(n)=3n$$

The first line in the solution says that $f$ is bijective. Why? I understand the rest of the solution, except this.

Answer 1

I found a proof showing that $f$ must be bijective, but it immediately solves the whole problem. I do not know if $0 \in \mathbb{N}$ in your notation, so I will assume $\mathbb{N} = \{1, 2, \ldots\}$. Note that $f(n) \geq 1$ for all $n \in \mathbb{N}$. Hence, $$ 3 \leq f(f(f(1))) + f(f(1)) + f(1) = 3. $$ So $f(1) = f(f(1)) = f(f(f(1))) = 1$. Inductively, suppose $f(k) = k$ for all $k \leq m - 1$. Then since Olivier Oloa already proved that $f$ is injective, we see that $f(n) \geq m$ for $n \geq m$. Hence, $$ 3m \leq f(f(f(m))) + f(f(m)) + f(m) = 3m, $$ showing that $f(m) = m$. Surjectivity follows, but also that $f$ is the identity function.