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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.

Mohsen Shahriari
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hoento
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1 Answers1

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As kindly explained below strictly increasing means $f(a) > f(b)$ for $a >b.$ Let's consider $f(0)$.

  1. $f(0) = 0$ is not possible because then $f^2(0) = 0 \neq 2.$
  2. $f(0) = 1$ is possible.
  3. $f(0) = m > 1$ is not possible, since $f^2(0) = f(m),$ and since $f$ is strictly increasing we have $f(m) > f(0) > 1$ which implies $f(m) \ge m + 2 > 2$ 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 $f(k+1)$?

Kore-N
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