Greatest prime divisor of consecutive integers Let $p,q$ be prime numbers with $p<q<2p$. Prove that there exists a pair of consecutive positive integers such that the greatest prime divisor of one of them is $p$ and the greater prime divisor of the other is $q$.
Such a prime number $q$ exists for any $p$ by Bertrand's postulate. For $p=3$ and $q=5$, we can take $5$ and $6$. For $p=5$ and $q=7$, we can take $14$ and $15$. I doubt there exists a general closed form in terms of $p$ and $q$ though.
 A: Bezout's theorem guarantees the existence of integers $m$ and $n$ satisfying
$$pm-qn=1$$
with $0\lt m\lt q$ and $0\lt n\lt p$. (Note, $m$ and $n$ are strictly positive since $p$ and $q$ are distinct primes.)  Letting $M=q-m$ and $N=p-n$ we also have
$$qN-pM=1$$
with $0\lt M\lt q$ and $0\lt N\lt p$.  Note that $n,N\lt p$ implies $q$ is the largest prime divisor of both $qn$ and $qN$ since $p\lt q$.  As for $m$ and $M=q-m$, they can't both be larger than $p$ since that would imply $q=m+M\gt2p$.  Thus one of them, at least, is less than or equal to $p$ and hence $p$ is the largest prime divisor of either $pm$ or $pM$ (or both).
As an example, let $p=47$ and $q=89$.  We have
$$47\cdot53-89\cdot28=1$$
which is not what we want, since $53$ is a prime larger than $47$, but we also have
$$89\cdot19-47\cdot36=1$$
which does give us what we want.
Remark:  It seems likely (or at least plausible) that for "most" pairs $(p,q)$, $p$ is the largest prime divisor of both $pm$ and $pM$.  In looking for an example where it isn't, I got lucky:  $(47,89)$ was the first pair I thought to try, and it produced $m=53$.
A: Partial answer:
case 1:
$x = a p$ 
$x + 1 = b q$
subtract
$1 =  b q - a p$
which is solvable for $(a,b)$ when gcd($p$,$q$) = $1$ by the Extended Euclidean Algorithm.
$b < p$ and $a < q$ this can be demonstrated by:
$1 \equiv  - a p$ (mod $q$) , $q$ is prime so an $a < q$ exists as an inverse to $-p$.
$1 \equiv  b q $ (mod $p$) , similarly $b < p$
Since $b < p < q$ then $q$ is the largest prime factor of $x+1$.
$a < q < 2p$  
Can't complete the proof that $p$ is the largest factor of $x$.

case 2:
$x = a q$ 
$x + 1 = b p$
subtract
$1 =  b p - a q$
Similarly: $b < q$ and $a < p$
Since $a < p < q$ then $q$ is the largest factor of $x$.
$b < q < 2p$
Can't complete the proof that $p$ is the largest factor of $x+1$.

A closed form for $x$:
case 1:
$qx = apq$
$p(x+1) = bpq$
Add or subtract:
$p(x+1) \pm qx \equiv 0$ (mod $pq$)
$x(p \pm q) \equiv -p$ (mod $pq$)
$x \equiv -p(p \pm q)^{-1}$ (mod $pq$)
case 2:
$x \equiv -q(q \pm p)^{-1}$ (mod $pq$)
