<s>[INST] Show that $\left[\alpha\right]+\left[\alpha+\frac{1}{m}\right]+\cdots + \left[\alpha+\frac{m-1}{m}\right] = \left[m\alpha\right]$ Let $m$ be a positive integer and let $\alpha$ be a real number. Show that
$$\left[\alpha\right]+\left[\alpha+\frac{1}{m}\right]+\cdots + \left[\alpha+\frac{m-1}{m}\right] = \left[m\alpha\right]$$ where $\left[\ \ \right]$ is the integral part function.
Remark: Attempt to prove this result by induction but  I failed in the inductive step, I have been unable use the induction hypothesis.
 [/INST] This is known as Hermite's identity. A nice proof is due to Matsuoka "On a Proof of Hermite's Identity", AMM 71:10 (1964), pp. 1115, DOI: 10.2307/2311413.
Let:
$\begin{align}
   f(x)
      = \lfloor m x \rfloor
   - \lfloor x \rfloor
   - \left\lfloor x + \frac{1}{m} \right\rfloor
   - \left\lfloor x + \frac{2}{m} \right\rfloor
   - \dotsb
   - \left\lfloor x + \frac{m - 1}{m} \right\rfloor
 \end{align}$
For real $\alpha, \beta$ it is $\lfloor \alpha + 1 \rfloor - \lfloor \beta + 1 \rfloor
      = \lfloor \alpha \rfloor - \lfloor \beta \rfloor$, so that:
$\begin{align}
    f\left( x + \frac{1}{m} \right)
      &= \lfloor m x + 1 \rfloor
    - \left\lfloor x + \frac{1}{m}\right\rfloor
    - \left\lfloor x + \frac{2}{m} \right\rfloor
    - \dotsb
    - \left\lfloor x + \frac{m - 1}{m} \right\rfloor
    - \lfloor x + 1 \rfloor \\
      &= \lfloor m x \rfloor
    - \lfloor x \rfloor
    - \left\lfloor x + \frac{1}{m} \right\rfloor
    - \left\lfloor x + \frac{2}{m} \right\rfloor
    - \dotsb
    - \left\lfloor x + \frac{m - 1}{m} \right\rfloor \\
      &= f(x)
  \end{align}$
On the other hand, for $0 \le x < 1 /m$ it is $f(x) = 0$, and so $f(x) = 0$ for all $x$, as was to be proved.
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