Sum and product of coprimes?

Question

Let's say $m_{1}, m_{2}, m_{3},...,, m_{\phi (m)}$ are the numbers which are coprime to $m$.

$m_{i}< m$ , $i=1,2,..,\phi (m)$, $\phi (m)$ is the Euler's totient function.

How to find sum and product of this numbers?

$$m_{1}+m_{2}+m_{3}+ \ldots +m_{\phi (m)}=?$$ $$m_{1}m_{2}m_{3} \ldots m_{\phi (m)}=?$$

Of course we have to use the prime divisors of $m$ but i couldn't find a proper way.

Answer 1

Here's a formula to find the product :

$$\prod_{I \subset [n] } \left (P_I^{\frac{m}{P_I}} \left(\frac{m}{P_I} \right )! \right)^{(-1)^{|I|}}$$

where $[n]=\{1,2,\ldots,n \}$ and $$P_I=\prod_{k \in I} p_k$$ where the prime factors of $m$ are $p_1,p_2,\ldots,p_n$ .

You can find it by a simple application of the Principle of Inclusion and Exclusion .

Answer 2

If $0 < a < m $ is coprime to $m$, then so is $m-a$. Thus $\sum_{(a,m) = 1} a = \frac{\varphi(m)}{2} m.$