TITLE: How find the $f(m+1)-f(m)=\text{constant},\forall m\in\Bbb{N}^{+}$
QUESTION [6 upvotes]: Let $f:\mathbb{N}^{+}\to \mathbb{N}^{+}$ be a strictly monotone increasing function such that
$$f(f(m+1))-f(f(m))=f(f(m+1)+1)-f(f(m)+1),\forall m\in \mathbb{N}^{+}.$$
Show that:
$$f(m+1)-f(m)=\text{ constant},\forall m\in \mathbb{N}^{+}$$
My attempt: Let $g(m)=f(f(m))-f(f(m)+1)$, then the condtion is 
$g(m+1)=g(m)$, so $g(m)$ is constant, or 
$f(f(m))-f(f(m)+1)$ be constant, then how prove it?

REPLY [5 votes]: Here is a counter-example:
Define $f:\mathbb{N}^+\rightarrow\mathbb{N}^+$ by the following rule
$$f(1) = 1,$$$$f(2) = 3,$$$$ f(k) = 2k-2, \forall k\geq 3.$$
Then, $\forall k \geq 3$,
$f(f(k+1))-f(f(k)) = f(2k) - f(2k-2) = 4k-2-(4k-6) =4\\
f(f(k+1)+1) - f(f(k)+1)=f(2k+1)-f(2k-1)=4k-(4k-4)=4.$
So the stated condition holds $\forall k \geq3$.
It's easy to check that it also holds for $k=1,2$.
Furthermore, $f$ is strictly monotone. Hence satisfies the statement.
However,$$ f(3)-f(2) = 1 \neq 2 = f(2)-f(1).$$