Solve $\frac{1}{2} n \varphi(n) \bmod n$

Question

How can I solve the equation \begin{align*} (a_1 + a_2 + \dots + a_{\varphi(n)}) \bmod n, \end{align*} where $a_1, a_2, \dots, a_{\varphi(n)}$ are the positive numbers relatively prime to $n$ ($n \in \mathbb{N}$), and smaller than $n$. I know that you can write this sum on the form \begin{align*} (a_1 + a_2 + \dots + a_{\varphi(n)}) = \frac{1}{2} n \varphi(n) \end{align*} and thus the equation simplifies to \begin{align*} \frac{1}{2} n \varphi(n) \bmod n \end{align*} I know that $n \bmod n = 0$. Does this mean that also $\frac{1}{2} n \varphi(n) \bmod n = 0$?

Answer 1

Use the fact that the Euler Totient function is always even for $n \ge 3$, so we must have that $\frac{\phi(n)}{2} \in \mathbb{Z}$, so therefore we have that $\frac 12 \phi(n)n$ is a mulitple of $n$.