Number theory related to rings

Question

Why it is true that every odd number r that is not equivlant to p mod p has a odd number $s$ s.t

$rs\equiv 2p-1 \mod 2p $

My first thought is that i should have an inverse for $r$ in the ring so i can write. $s\equiv r^{-1} (-1) \mod 2p $ but im not sure what in number theory tells me that $r^{-1} $ exists?

Where p as always in number theory is prime.

Answer 1

Assuming $p$ is a prime, and that not just $r\ne p$ but $p\not\mid r$...

$r^{-1}$ exists because $r$ (being odd) is coprime with $2p$. Thus, $1=rs+2pt$ for some $s,t\in\mathbb Z$ (Bézout's identity), or, reducing modulo $2p$, $1\equiv rs\pmod{2p}$.

Answer 2

Indeed, we can take $-1/r=-r^{-1}$ in $\mathbb{Z}/2p$, since we know that $r\not\equiv 0\bmod 2$ and $r\not\equiv 0\bmod p$ by assumption.