Question: 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?

Answer: 
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,$$ f(3)-f(2) = 1 \neq 2 = f(2)-f(1).$$
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 $m$.