Positive integer solution $pm = qn+1$ 
Let $m,n$ be relatively prime positive integers. Prove that there exist positive integers $p,q$ such that $pm = qn+1$.

We know Bézout's identity that there exist integers $p,q$ such that $pm+qn = 1$, but how do we know we can get positive integers $p,q$ with $pm = qn+1$?
 A: Clearly $p$ and $q$ cannot both be negative in your Bézout equation. Also note that you can "shuffle" around things.
If $(p',q')$ is a solution of the Bézout equation then so is $(p'+n, q'-m)$. In this way you can make sure you have a solution $(p'',q'') $ where the coefficient $q''$ of $n$ in the Bézout identity is negative (and the other is positive). 
Now, just move the $q''n$ to the other side to get $p''m = -q''n +1$, where $p''$ is positive and $q''$ negative, so $-q''$ positive.  
A: (1). Given integers $p',q'$ with $p'm+q'n=1,$ the set of all $(p'',q'')$ such that $p''m+q''n=1$  is $\{(p'+xn, q'-xm): x\in Z\}.$ 
If $p'>0$ then $q'<0,$ so let $p=p'$ and $q=-q'.$
If $p'\leq 0$ take $x\in Z^+$ where $x$ is large enough that $p'+xn>0  $ and $ xm-q'>0 . $ Let $p=p'+xn$ and $q=-q'+xm.$  
(2). A one-step general way is that $$p'm\equiv 1\pmod n\implies \forall x\in Z\;((p'+xn)m\equiv 1\pmod n).$$  So for $x$ positive and large enough that $p'+xn>0$ we have $0<(p'+xn)m=1+qn$ for some $q\in Z,$ so $q>0.$
