How find the $f(m+1)-f(m)=\text{constant},\forall m\in\Bbb{N}^{+}$ 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?
 A: 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,$$ f(3)-f(2) = 1 \neq 2 = f(2)-f(1).$$
A: So we are to prove that, given $f:\;N^ +   \to N^ +  $, we have
$$ \bbox[lightyellow] {  
\eqalign{
  & f\left( {f\left( {m + 1} \right)} \right) - f\left( {f\left( m \right)} \right) = f\left( {f\left( {m + 1} \right) + 1} \right) - f\left( {f\left( m \right) + 1} \right)\quad \left| {\;\forall m \in N^ +  } \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad f\left( {m + 1} \right) - f\left( m \right) = \Delta \,f\left( m \right) = const\quad \left| {\;\forall m \in N^ +  } \right. \cr} 
 }$$
which means
$$ \bbox[lightyellow] {  
\eqalign{
  & f\left( {f\left( {m + 1} \right)} \right) - f\left( {f\left( m \right)} \right) = f\left( {f\left( {m + 1} \right) + 1} \right) - f\left( {f\left( m \right) + 1} \right)  \cr 
  & \quad \quad  \Downarrow   \cr 
  & f\left( {f\left( m \right) + 1} \right) - f\left( {f\left( m \right)} \right) = f\left( {f\left( {m + 1} \right) + 1} \right) - f\left( {f\left( {m + 1} \right)} \right)  \cr 
  & \quad \quad  \Downarrow   \cr 
  & \Delta \,f\left( {f(m)} \right) = \Delta \,f\left( {f(m + 1)} \right) = \Delta \,f\left( {f(m) + \Delta \,f(m)} \right) \cr} 
 } \tag{1}$$
In the post it is further specified that $f$ is
a strictly monotone increasing function.
and according the  definition of monotonic function as per Wikipedia
 a) either the monotone specification is superfluous, and the function is just strictly increasing,
 b) or it is actually meant that it is strictly increasing and with monotone increase.
So let's examine both cases.
a) strictly increasing
The image of $f(m)$ will not coincide, in general, with its domain. So from (1) we cannot deduce the claim
in general and therefore we shall reject the hypothesis.
For example, a function like this one
$$
\begin{array}{c|ccccc}
   m  &  1 & 2 & 3 & 4 &  \cdots   \\
\hline
   {f(m)}  &  1 & 3 & 6 & 8 &  \cdots   \\
\end{array}
$$
gives
$$
\begin{array}{l}
 f(f(2)) - f(f(1)) = f(3) - f(1) = 5 =  \\ 
  = f(f(2) + 1) - f(f(1) + 1) = f(4) - f(2) = 5 \\ 
 \end{array}
$$
but
$$
f(2) - f(1) = 2\quad  \ne \quad f(3) - f(2) = 3
$$
b) strictly increasing and with monotone increase
In this interpretation it is meant that
$$ \bbox[lightyellow] {  
\eqalign{
  & m_{\,1}  < m_{\,2} \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  f(m_{\,1} ) < f(m_{\,2} ) \hfill \cr 
  \left( {\Delta \,f(m_{\,1} ) \le \Delta \,f(m_{\,2} )} \right)\; \hfill \cr}  \right.\quad  \vee \quad \left\{ \matrix{
  f(m_{\,1} ) < f(m_{\,2} ) \hfill \cr 
  \left( {\Delta \,f(m_{\,1} ) \ge \Delta \,f(m_{\,2} )} \right) \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left( {0 < \Delta \,f(m_{\,1} ) \le \Delta \,f(m_{\,2} )} \right)\;\quad  \vee \quad \Delta \,f(m_{\,1} ) \ge \Delta \,f(m_{\,2} ) > 0 \cr} 
 } \tag{2}$$
This tells us that $\Delta f$ is either non-decreasing or non-increasing, and identity (1) that 
in various points it is stable, thus  in this acception the claim  follows immediately.
A: BIG FAT EDIT:
I will build a counterexample to the problem. The problem is to prove that a function satisfying $f(f(m)+1)-f(f(m))=constant=K   (A)$ implies $f(m+1)-f(m)=constant=C   (B)$ because the original statemtn is equivalent to (A).
Lets start building the function and see what intuition we can gain in the process. Once we determine a value for 1, this is $f(1)$ shall we call it $a$ then if a>2 we have all the values till a that are determined by nothing, but once we put a alue to a we get a value for a+1. So we need to know which values of the naturals we are giving to the function and register them in order to remember when we will need to satisfy (A).
Lets say $a=3$ and also $K=3$ $f(2)=5$ and $f(3)=7$, I assigned those because they were free, but $f(4)=10$ yet we dont mind if $f(5)=12$ because this doesnt exclude $f(6)-f(5)=3$ (I care about 5 because it belongs to the span of the function) so we get the thing here. We only need to give consistent values to the span of f so I conserve the necessary properties.
So, we register the span (3,5,7)once in $x=3$ we jump $y=3$ to make $f(4)=10$ (and add 10 to the span (3,5,7,10))but because 4 didnt beling to the range previously registered we dont need $f(4)+3=f(5)$ so to prove how great we are we actually say $f(5)=f(4)+2=12$ (and add 12, (3,5,7,10,12)) and then we can build a function so that in these new added "spans" we "jump" 3 in the y axis to preserve the property, (also happy that this preserves monotonicity) AND we can actually violate this for other naturals in the domain. Whenever we are not on the "spans" we only add 2 to the previous value creating a slope 2 line and we add a new span, when we get to one span we jump 3 to preserve the property. 
This produces a broken line so the conclusion doesnt follow. 
A: It seems (there are 4 a.m. here and I’m tired and sleepy now, so I may have hallucinations :-) ) that there indeed is a counterexample. Put $f(m)=m^2$ if $m$ is even and $f(m)=(m-1)^2+1$, if $m$ is odd. Then $f(m+1)-f(m)$ equals $1$ for even $m$ but equals $4m-1$ for odd $m$, so the function $f$ is strictly monotone increasing, but $f(m+1)-f(m)$ is not constant. It remains to remark that $f(m)$ is even for each $m$, so $$f(f(m+1))-f(f(m))-f(f(m+1)+1)+f(f(m)+1)=$$ $$f(m+1)^2-f(m)^2-f(m+1)^2-1+f(m)^2+1=0.$$
PS. An even simpler counterexample we obtain if we put $f(m)=2m$ if $m$ is even and $f(m)=2m+1$, if $m$ is odd. Then $f(m+1)-f(m)$ equals $3$ for even $m$ but equals $1$ for odd $m$. 
