If $m,n$ are natural non-zero numbers show that $$[x]+[x+1/n]+[x+2/n]+...+[x+m/n]=[nx]$$ for any real $x$ if and only if $m=n-1$. $[x]$ is the integer part of $x$.
I know from the Hermite Identity that if $m=n-1$ then $$[x]+[x+1/n]+[x+2/n]+...+[x+(n-1)/n]=[nx]$$ but I do not know how to prove that if $$[x]+[x+1/n]+[x+2/n]+...+[x+m/n]=[nx]$$ then $m=n-1$. Should I also use the Hermite Identity for this?
$\newcommand{\fl}[1]{\left\lfloor #1 \right\rfloor}$Fix an integer $m \geqslant 1$ and define $$f(x) = -\fl{nx} + \sum_{k = 0}^{m} \fl{x + \frac{k}{n}}.$$ Suppose that $f(x) = 0$ for all $x \in \Bbb R$.
We wish to show that $m = n - 1$.
First, note that $f(0) = 0$ immediately tells us that $m \leqslant n - 1$.
For the sake of contradiction, assume that $m \leqslant n - 2$. Then, we have $$f\left(\frac{1}{n}\right) = -1 + \sum_{k = 0}^{m} \fl{\frac{k + 1}{n}} = -1,$$ since every term in the summation is $0$. But this means that $f(1/n) \neq 0$, the desired contradiction.