Find all functions $f$ that maps from $\mathbb{N}$ to $\mathbb{N}$, and $$f(n) + f(f(n)) = 2n + 3$$ for all $n \in \mathbb{N}$
Attempt:
I find a function that is $f_{1}(n) = n+1$, we can easily check $ f_{1}(f_{1}(n)) = n + 2 $, so adding together we always get $2n+3$. So in this case they are consecutive positive integers.
In general for linear function: $f(n) = An +B$, then $f(f(n)) = A(An+B) + B = A^{2}n + AB + B$, so
$$ f(n) + f(f(n)) = (A + A^{2})n + AB+2B $$
if $A,B$ integers, then the only solution is $A=B=1$.
Another fact is that $f(n) + f(f(n))$ is odd for all natural numbers $n$. This means that $f(n)$ and $f(f(n))$ must have different parity.
I will add something new:
$$ f(f(n)) + f(f(f(n))) = 2f(n)+3 $$
$$ f(f(f(n))) + f(f(f(f(n)))) = 2f(f(n))+3 $$
$$ \vdots $$
$$ F(n, k) + F(n, k+1) = 2F(n, k-1) + 3$$
where $F(n,k) = \underbrace{f(f(f(...(f(n))...)))}_{k}$ if $k \ge 1$, and $F(n, 0) = n$.