Sum of elements in reduced residue system modulo n is divisible by n

Question

Prove that sum of elements in reduced residue system modulo $n \in N$ is divisible by $n$.

I feel like problem just comes down to pairing elements of RRS in way that they are congruent, but can't quite work it out. Please post detailed and readable answer, as most of problems I need to do basically comes down to this one ..

I kind of feel that the accepted answer in that post is more math involved. If you feel the answer and hints provided there are sufficient, then you may delete the post. — Maged Saeed, Dec 18 '18 at 18:42

score 1 · Accepted Answer · answered Dec 18 '18 at 18:18

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The reduced residue system of $2$ is only $1$. But $2\not | 1$, so, $n > 2$.

Let us address this problem for numbers greater than $2$. Now, since $\gcd(a,n) = 1$, what can you say about $\gcd(n-a,n)$? It is also $1$. Now, we can use this to pair elements of the Reduced Residue System, as you have suggested, as follows:

$a_1+a_2+a_3+\cdots+a_{\phi(n)\over2}+(n-a_1)+(n-a_2)+\cdots+(n-a_{\phi(n)\over2}) = \frac{\phi(n)}{2}n$. This also can give you a hint why $2$ is a counter example as $\phi(n)$ is only odd for $n=2$. This answer is collected from answers in this post.

answered Dec 18 '18 at 18:18

Maged Saeed

1,140

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Ooh, right .. So if $k \in RRS$ then $n-k \in RRS$ because $(n-k, n) = 1$ and $n-k \ne k mod n$ ? – Dec 18 '18 at 18:32
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Yeah, you got it. – Maged Saeed Dec 18 '18 at 18:37

Sum of elements in reduced residue system modulo n is divisible by n

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