Find all functions $f:\mathbb{Z} \to \mathbb{Z} $ such that$f(0)=1$ and $$f(f(n))=f(f(n+2)+2)=n. \quad \forall n \in \mathbb{Z} $$

My approach: Plugging in some values, it is not hard to see that $f(n)=1-n$ satisfies the given relation. I claim that $ f(k)=1-k $ for some $k \in \mathbb{Z}$.

I just cannot see a way to use the relation and induct on $k$ to prove my hypothesis. Am I missing something obvious? Please help as I am new to functional equations. Also please share some online resources to solve functional equations as I am preparing for Olympiads.

Thank you.

  • $\begingroup$ "I claim that $ f(k)=1-k$ for some $k \in \mathbb{Z}$ ": you just said that it holds for all $k$ ! $\endgroup$ – Yves Daoust Feb 12 '18 at 9:25
  • $\begingroup$ @YvesDaoust shouldn't I do that for some k to proceed with my induction hypothesis? $\endgroup$ – alphasquared Feb 12 '18 at 13:20
  • $\begingroup$ You say "it is not hard to see that $f(n)=1−n$ satisfies the given relation", don't you ? $\endgroup$ – Yves Daoust Feb 12 '18 at 13:37
  • $\begingroup$ @YvesDaoust I'm sorry I didn't mean that. i actually meant to point out what i thought the function according to me is, to the viewers of this question. $\endgroup$ – alphasquared Feb 12 '18 at 13:47

$f(f(n)) = n \ (\forall n \in \mathbb{Z})$ implies $f$ is injective, thus$$ f(f(n)) = f(f(n + 2) + 2) \Longrightarrow f(n) = f(n + 2) + 2. \quad \forall n \in \mathbb{Z} $$ Also$$ 0 = f(f(0)) = f(1), $$ then$$ f(n) = -n + 1 \quad (\forall n \in \mathbb{Z}) $$ can be proved by induction.

  • $\begingroup$ How do I proceed with the inductive step? $\endgroup$ – alphasquared Feb 13 '18 at 10:05
  • 1
    $\begingroup$ First using $n = 0, 1$ as base cases to prove for $n \geqslant 0$, then using $n = 0, 1$ as base cases to prove for $n < 0$. $\endgroup$ – Saad Feb 13 '18 at 10:13
  • $\begingroup$ Please check this: suppose that $f(k)=1-k$ for some $k \in \mathbb{Z}$. Then we have $f(k)=f(k+2)+2$ which implies $f(k) > f(k+2)$ which implies $f(k) \geqslant f(k+1)$. But $f(k) = f(k+1)$ implies $0=1$ which is absurd. Therefore $f(k) > f(k+1)$ thus forcing $f(k+1) = -k$. (It can be seen using the relation that $f(k+2)= -1-k$) $\endgroup$ – alphasquared Feb 13 '18 at 10:23
  • $\begingroup$ No need to be so complicated since $f$ has the recurrence relation $f(n + 2) = f(n) - 2$ with initial conditions $f(0) = 1$, $f(1) = 0$ for $n \in \mathbb{N}$. $\endgroup$ – Saad Feb 13 '18 at 10:59
  • $\begingroup$ That $+2$ inside the recurrence function is making it difficult for me to induct on consecutive integers, instead I am ending up inducting on consecutive integers with same parity. $\endgroup$ – alphasquared Feb 13 '18 at 11:04

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