I have been looking for theorems that say something about relations of factors between consecutive numbers. With Bezout's identity it can be proved that if $\gcd(a,b)=1$ there is always a number $n$ such that $a|n$ and $b|(n+1)$, which tells something about independency between factors of consecutive numbers.
After a lot of work I found a simple formula that seems to determine the least non-negative number $n$ as above, $n=b\cdot r^{-1}-1$, where $r^{-1}$ is the number corresponding to the inverse of $r$ in the multiplicative group of numbers coprime to $a$ and $r$ is the reminder when $b$ is divided with $a$.
Obviously $b|(n+1)$ but how to prove that $a|n$ and that there are no smaller number than $n$?
Given that $n+1 = b r^{-1}$ and $b \equiv r \mod a$ it follows that $br^{-1} = 1 \mod a$ and $ br^{-1} -1 \equiv 0 \mod a$ means that $n \equiv 0 \mod a$. So $a \mid n$.
Suppose $m$ is another positive integer such that $a \mid m$ and $b \mid m+1$.
So there exist $k$ such that $m+1 = kb$. So $$ 1\mod a \equiv (m+1) \mod a \equiv (kb) \mod a \equiv (k \mod a)(b \mod a) \equiv r(k \mod a)$$
So we have $$k \equiv r^{-1} \mod a$$
The least value $k$ satisfying the last equation is $r^{-1}$ so $k \ge r^{-1}$. This implies that $m = kb -1 \ge r^{-1}b -1 = n$.