TITLE: How to prove that gcd(m+1, n+1) divides (mn-1)
QUESTION [9 upvotes]: I'm learning about divisors and the gcd, but now I'm stuck at proving:

gcd(m+1, n+1) divides (mn-1) for all m,n in the set of Integers

Help is appreciated on how to prove this!
Thanks

REPLY [5 votes]: Although it's less elegant than the other approaches, you can also prove this through direct substitution. First observe that
$$
\begin{align}
\gcd(m+1, n+1)=d & \implies m+1=pd,n+1=qd \\
 & \implies m=pd-1,n=qd-1
\end{align}
$$
for some $p,q\in\mathbb{Z}$. Then
$$
\begin{align}
mn-1 & = (pd-1)(qd-1)-1 \\
 & = (pqd^2-(p+q)d+1)-1 \\
 & = (pqd-(p+q))d,
\end{align}
$$
which is an integer multiple of $d$.