How to prove $n$ is prime? Let $n \gt 1$ and 
$$\left\lfloor\frac n 1\right\rfloor + \left\lfloor\frac n2\right\rfloor + \ldots + \left\lfloor\frac n n\right\rfloor = \left\lfloor\frac{n-1}{1}\right\rfloor + \left\lfloor\frac{n-1}{2}\right\rfloor + \ldots + \left\lfloor\frac{n-1}{n-1}\right\rfloor + 2$$ 
and $\lfloor \cdot \rfloor$ is the floor function. How to prove that $n$ is a prime?
Thanks in advance.
 A: i have a question if i define the function $ \sigma (x)= \sum_{n=1}^{x}d $ and 'd' is the function divisor isn't the problem $ \sigma(x)=\sigma(x-1)+2 $
A: You know that 
$$\left( \left\lfloor\frac n 1\right\rfloor - \left\lfloor\frac{n-1}{1}\right\rfloor \right)+\left(  \left\lfloor\frac n2\right\rfloor - \left\lfloor\frac{n-1}{2}\right\rfloor\right) + \ldots + \left( \left\lfloor\frac{n}{n-1}\right\rfloor - \left\lfloor\frac{n-1}{n-1}\right\rfloor\right) + \left\lfloor\frac n n\right\rfloor=+2 \,.$$
You know that
$$\left( \left\lfloor\frac n 1\right\rfloor - \left\lfloor\frac{n-1}{1}\right\rfloor \right)=1$$
$$\left\lfloor\frac n n\right\rfloor =1$$
$$\left(  \left\lfloor\frac n k\right\rfloor - \left\lfloor\frac{n-1}{k}\right\rfloor\right) \geq 0, \qquad \forall 2 \leq k \leq n-1 \,.$$
Since they add to 2, the last ones must be equal, thus for all $2 \leq k \leq n-1$ we have
$$ \left\lfloor\frac n k\right\rfloor - \left\lfloor\frac{n-1}{k}\right\rfloor = 0 \Rightarrow  \left\lfloor\frac n k\right\rfloor = \left\lfloor\frac{n-1}{k}\right\rfloor $$
It is easy to prove that this means that $k \nmid n$. Since this is true for all $2 \leq k \leq n-1$, you are done.
