Let $f$ be a bijection from $\mathbb N$ to itself. Prove that one can always find three natural number $a,b,c$ such that $a<b<c$ and $f(a)+f(c)=2f(b)$.
Source: French MO 1995 P5
Tags: function, algebra
Let $f$ be a bijection from $\mathbb N$ to itself. Prove that one can always find three natural number $a,b,c$ such that $a<b<c$ and $f(a)+f(c)=2f(b)$.