$\!\bmod n\!:\,$ a complete residue system has sum $\equiv 0$ if $n$ is odd, else $\equiv \frac{n}{2}$

Question

I have came across this mathematical question, and it has been mind-provoking me, I hope to find some help here.

The question is as follows: Given an SDR (System of Distinct Representatives) set ($a_0,a_1,\ldots,a_n$) in relation to modulo $n$. Let $s=\sum_{i=1}^{n} a_i$. a.) prove that $s=0 \pmod n$ if $n$ is odd. b.) prove that $s=\frac{n}{2} \pmod n$ if $n$ is even.

Thanks!

Answer 1

Hint $\ $ The set is closed under negation, whose non-fixed points $\rm\ {-}k\not\equiv k\:$ pair up and contribute zero to the sum, leaving only the sum of the $\rm\color{#0a0}{fixed\ points}$, where $\rm\, \color{#0a0}{-k\equiv k} \iff 2k\equiv 0,\, $ therefore $\rm\ k\equiv 0\ $ if $\rm\: n\:$ is odd, $ $ else $\rm\ k \equiv 0,\ n/2.$

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Remark $\ $ This is a special case of Wilson's theorem for groups - see my answer here - which highlights the key role played by symmetry (here a negation reflection / involution).

See also Gauss's grade-school trick for summing an arithmetic progression.