Show that if $c_1, c_2, \ldots, c_{\phi(m)}$ is a reduced residue system modulo m, $m \neq 2$ then $c_1 + \cdots+ c_{\phi(m)} \equiv 0 \pmod{m}$ 
Show that if $c_1, c_2,\ldots, c_{\phi(m)}$ is a reduced residue system modulo $m$, $m \neq 2$, and $m$ is a positive integer, then $c_1 +\cdots+ c_{\phi(m)} \equiv 0 \pmod{m}$

From the problem statement, I only know that $\gcd(c_i, m ) = 1$.
Is there any related theorem that I missed?
A hint would be greatly appreciated.
Thanks,
Chan
 A: HINT: If $c_i$ is a reduced residue class, then so is $m-c_i$. (Why?) and $\phi(m)$ is even $\forall m >2$
A: Hint $\ $ It's 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). 
Said more simply, since your set is closed under the negation reflection, its non-fixed points $\rm -k\not\equiv k\:$ pair up and contribute zero to the sum, leaving only the sum of its fixed points $\rm - k\equiv k \iff 2k\equiv 0,\ $ so $\rm\ k\equiv 0\ $ if $\rm\: m\:$ is odd, else $\rm\ k \equiv 0,\ m/2$.
See also Gauss's grade-school trick for summing an arithmetic progression.
A: Clearly $g.c.d(m-1,m) = 1$. Then $\{(m-1)c_{1},(m-1)c_{2},...,(m-1)c_{\phi(m)}\}$ are all relatively prime to m; furthermore, they are mutually incongruent, since $(m-1)c_{i} \equiv (m-1)c_{j} \ \ (mod \ m)$ implies that $c_{i} \equiv c_{j} \ \ (mod \ m)$, by the cancellation law. We may thus pair each $(m-1)c_{i}$ with some $c_{j}$ such that $(m-1)c_{i} \equiv c_{j} \ \ (mod \ m)$, and we note that $c_{j}$ is uniquely defined for each $(m-1)c_{i}$. Then $$(m-1) c_{i} + (m-1)c_{w} \equiv c_{j} + (m-1)c_{w} \ \ (mod \ m) \equiv c_{j} + c_{k} \ \ (mod \ m) $$. Then $$\sum_{l=1}^{\phi(m)} (m-1) c_{l} \equiv \sum_{l=1}^{\phi(m)} c_{l} \ \ (mod \ m)$$. This implie $$2\sum_{l=1}^{\phi(m)} c_{l} \equiv m\sum_{l=1}^{\phi(m)} c_{l}  \ \ (mod \ m) \equiv 0 \ \ (mod \ m)$$. If $g.c.d(2,m) = 1$, that is, m is odd, we have $$\sum_{l=1}^{\phi(m)} c_{l} \equiv 0 \ \ (mod \ m)$$.
