1. ## primitive roots

Let n>1 be a positive integer. Find the number of all m, so gcd(n,m)=1,
$1\le m<n$ and $mn-1\equiv 1$ (mod n).

2. ## Re: primitive roots

This is because $mn-1 \equiv 1 \text{ (mod n)}$ if and only if $-1 \equiv 1 \text{ (mod n)}$. Note that $mn$ is a multiple of $n$ for any integers $m,n$, so that factor is equivalent to $0\text{ (mod n)}$.
$\displaystyle mn-1=1 \pmod{n} \implies mn=2 \pmod{n} \implies 0=2 \pmod{n}$