Let $f$ be a bijection from the set of non-negative integers to itself. Show that there exist integers $a$,$b$,$c$ such that $a < b < c$ and $f(a)+f(c)=2f(b)$.
I don't know how to approach this problem. I think I'll have to show some kind of contradiction. But, I haven't managed to do anything fruitful yet. Any help would be appreciated.
We take a such that $f(a) = 0$.
Ad absurdum, we suppose : $\forall b>a, \forall c >b, f(c) \neq 2f(b)$ (*)
As f is a bijection, $2f(b) \in f(\mathbb{N})$, but $f(b) \neq 0$ so $f(b) \neq 2 f(b)$, and (*). Hence $2f(b) \notin f([\mid b, +\infty \mid ])$. Therefore :
$\forall b>a, \exists c < b, f(c) = 2f(b)$. By induction it is then clear that $\forall b>a, \exists k>0, \exists C \leqslant a, f(C) = 2^k f(b)$ (base case : a+1 ; inductive step on b).
From that you can deduce quickly a contradiction, using that f is a bijection (so $\{f(0),..,f(a)\}$ should be an infinite set).
We conclude that : $\exists b>a, \exists c>b, 0 + f(c) = 2f(b)$.