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

Solution 1:

HINT: If $c_i$ is a reduced residue class, then so is $m-c_i$. (Why?) and $\phi(m)$ is even $\forall m >2$

Solution 2:

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.