Rijul saini wrote: Indian Postal Training Set 3, Problem 7 Does there exist a function $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for every $n \ge 2$, \[f (f (n - 1)) = f (n + 1) - f (n)?\] see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=376900

Note that $f(n+1)>f(n)$ for all $n\ge2$, so $f$ is increasing in $[2,\infty)$. We get $f(n)\ge n-1$. Fix $n\ge2010$, \[\sum_{i=2}^{n+1}f(f(i-1))=\sum_{i=2}^{n+1}(f(i+1)-f(i))\] \[f(f(1))+f(f(2))+\ldots+f(f(n))=f(n+2)-f(2)\] So $f(n+2)>f(f(n))$, which gives $n+2>f(n)$, and analogously $n+4>f(n+2)$. We also have \[f(n+2)>f(f(2))+\ldots+f(f(n))\ge1+2+\ldots+n-1=\frac{n(n-1)}2\] So $n+4>\frac{n(n-1)}2$, which is obviously wrong.