Prove that for each odd $m> 2$, is true that $\displaystyle \sum_{k=1}^{m-1} k \equiv 0 \pmod m$

Question

please, know somebody solution for this argument?

Prove, that for each odd $m > 2$, it is true that $$\sum_{k=1}^{m-1} k \equiv 0 \pmod m$$

Thanks for yours answers!

Answer 1

$$\sum_{k=1}^{m-1} k = \frac{(m-1)m}{2}$$

Note that if $m$ is odd, then $m-1$ is an even integer, which means we can write $\frac{(m-1)m}{2} = am$ for some integer $a$. And $am \equiv 0 \bmod{m}$ by inspection.