Is sum of all positive integers less than and relatively prime to $n$ is divisible by n, where $n > 2$?

Question

We observe that the statement is true for $n$ a prime. But what happens when $n$ is composite? I have verified it for some numbers. However, I am not finding any proof or counter example. Can anyone help me with this?

Answer 1

Exactly, we can prove that

If $n>1$, then the sum of all positive integers less than $n$ and prime to $n$ is $\frac{n}{2}\phi(n)$, where $\phi$ is Euler-phi function.

Let $a$ be a positive integer less than $n$ and prime to $n$. Then $au+nv = 1$, for some $u,v \in \Bbb Z$. Therefore $n(v+u)+ (n-a)(-u) = 1$. Since $u+v, -u$ are integers, $n-a$ is also a positive integer less than $n$ and prime to $n$. Therefore $\phi(n)$ positive inegers less than $n$ and prime to $n$ can be expressed as $a_1,a_2,\cdots,(n-a_2),(n-a_1)$.

Let $S = a_1+a_2+\cdots+(n-a_2)+(n-a_1)$. Also, write it in a reverse order that is $S = (n-a_1)+(n-a_2)+\cdots+a_2+a_1$. Which implies $2S = \phi(n).n \iff S = \frac{n}{2}\phi(n)$.

Now observe only one thing: $\phi(n)$ is an even integer if $n>2$. For proof of this statement, you can look at this.