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

Question

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

Answer 1

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

Answer 2

Hint $ $ It is a special case of Wilson's theorem for groups - see my answer here - which highlights the key role played by symmetry (here a $\rm\color{#c00}{negation\ reflection}$ / involution).

Said simpler, since $\,k\,$ is coprime to $\,m\!\iff\! \color{#c00}{-k}\,$ is coprime to $m$, the residue system is closed under $\rm\color{#c00}{negation\ reflection}$, so non-fixed points $\, -k\not\equiv k\,$ all pair up and cancel out of the sum, leaving us with only the sum of the $ $ fixed points $\, -k\equiv k \!\iff\! 2k\equiv 0,\ $ so $\, k\equiv 0\ $ if $\rm\: m\:$ is odd, else $\,k \equiv 0,\ m/2$.

More concretely, every complete residue system is congruent to a balanced residue system, e.g. $\,0,\pm1, \pm 2,\pm3 \pmod{\!7}\,$ and $\,0,\pm1,\pm2, 3\pmod{\!6},\,$ whose sums are clearly $\,0\,$ or $\,m/2,\,$ with same sums for the subset of residues coprime to $\,m,\,$ due to said key property of coprimes being closed under negation.

See Gauss's grade-school trick for summing an arithmetic progression. for further examples and elaboration.

Answer 3

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 $$\sum_{l=1}^{\phi(m)} c_{l} \equiv 0 \ \ (mod \ m)$$.