Functional Equation Problem: Strictly Increasing $ f $ satisfying $ f \big( f ( n ) \big) = n + 2 $ If $ \mathbb N $ denotes all positive integers, then find all functions $ f : \mathbb N \to \mathbb N $ which are strictly increasing and such for all positive integers $ n $, we have:
$$ f \big( f ( n ) \big) = n + 2 $$

So far I know that $f(n)$ is greater than $n$ because the function is strictly increasing, but I'm not sure how to use this in order to solve the equation.
As kindly explained below strictly increasing means $f(a) &gt; f(b)$ for $a &gt;b.$
Let's consider $f(0)$.


$f(0) = 0$ is not possible because then $f^2(0) = 0 \neq 2.$
$f(0) = 1$ is possible.
$f(0) = m &gt; 1$ is not possible, since $f^2(0) = f(m),$ and since $f$ is strictly increasing we have $f(m) &gt; f(0) &gt; 1$ which implies $f(m) \ge m + 2 &gt; 2$ in contradiction with the assumption $f^2(0) = 2.$


Hence $f(0) = 1.$ Now suppose there is a $k$ such that $f(k) = k+1.$ What can you say about $f(k+1)$?
Hint: Set
$$f(n)=an + b$$
Viz,
\begin{align}
f(f(n)) &amp;= a(an+b)+b \\
&amp;= a^2n+ b(a+1)
\end{align}
Thus $a= \pm 1$ and
$$
 b(a+1) = \begin{cases}
      0, &amp; \text{if}\ a=-1 \\
      2b, &amp; \text{if}\ a= +1
    \end{cases}
$$