Show that if $c_1, c_2,\cdots, c_{\phi (m)}$ is a reduced residue system modulo m, where m is a positive integer with $m \neq 2$, then $c_1+c_2+\cdots+c_{\phi (m)}=0 modm$
Please, help

Question

ikram002p · Answer

so no two elements in   $c_1, c_2,\cdots, c_{\phi (m)}$  have same modulo of m but yet  each one have some modulo of m, right ?

loser66 · Answer

I can see it in small case, but don't know how to generalize it. See!!
if m =10, then reduced residue modulo 10 is {1,3,7, 9}, then 1+3+7+9 =20 and $20\equiv 0(mod 10)$
but how to argue with m case?

ikram002p · Answer

well read my first comment which means in some order we can rewrite  the set as :-
$c_1, c_2,\cdots, c_{\phi (m)} =(1,2,3..., m-1 \mod m)$

ikram002p · Answer

agree to this ?

loser66 · Answer

Nope :) since m is a positive integer, we don't know what it is yet, if it is an even number, then 2 is not in the set. Am I right?

ikram002p · Answer

wait sorry let me read it again =)

ganeshie8 · Answer

Hints : 

1) $\phi(m)$ is even for $m\gt 2$

2) If $c_1$ is in the residue set, then $m-c_1$ is also in the residue set.

loser66 · Answer

Got you @ganeshie8 and I got the answer also. :)