Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that $$f(n)+2f(f(n))=3n+5$$
In the book, the following solution is provided. For this post, only Case 1 is relevant.
From $f(1)+2f(f(1)) = 8$ we deduce that $f(1)$ is an even number between 1 and 6, that is, $f(1)=2, 4,$ or $6$.
$\textbf{Case 1:}$ If $f(1) = 2$ then $2+2f(2) = 8$, so $f(2) = 3$. Continuing with $3+2f(3) = 11$, we obtain $f(3)=4$, and formulate the conjecture that $f(n)=n+1$ for all $n \geq 1$. And indeed, in an inductive manner we see that $f(n)=n+1$ implies $n+1+2 f(n+1)=3 n+5$; hence $f(n+1)=n+2$
$\textbf{Case 2:}$ $f(1)=4$ gives $4+2 f(4)=8$, so $f(4)=2$. But then $2+2 f(f(4))=17$, which cannot hold for reasons of parity.
$\textbf{Case 3:}$ If $f(1)=6$, then $6+2 f(6)=8$, so $f(6)=1$. This cannot happen, because $f(6)+2 f(f(6))=1+2 \cdot 6$, which does not equal $3 \cdot 6+5$.
We conclude that $f(n)=n+1, n \geq 1$, is the unique solution to the functional equation.
$\textbf{My Doubt:}$ In Case 1, they claimed $f(n)=n+1$, and used induction to prove it. But how are they sure there doesn't exist other solutions? Induction doesn't prove that it's the only solution available with $f(1) = 2$ and $f(2) = 3$.
Best Answer
They don't show that $f(n)=n+1$ satisfies the conditions but rather that for each $n$, $f(n)$ HAS TO BE $n+1$. Here is how it works.
We have $f(1)=2$. That gives us \begin{align}2+2f(2)=\ \ 8&\implies f(2)=3\\ 3+2f(3)=11&\implies f(3)=4\\4+2f(4)=14&\implies f(4)=5\end{align} and so on. The point is that $f(n)=n+1$ for some $n$ forces $f(n+1)=n+2$. So since it is true for $n=1$, it has to be true for $n=2$, $n=3$ and so on for each $n$ inductively.
Hope this helps. Ask anything if not clear. :)