For what pairs of a and b is $a|b$ and $a+1|b+1$. where $|$ mean "is a divisors of".

 

There exist $p$ and $q$ so $ap=b$ and $(a+1)q = b+1$

Now is 

$$b-a=ap-a=a(p-1)$$

$$b-a=(b+1)-(a+1)=(a+1)(q-1)$$

so

$$a(p-1)=(a+1)(q-1)$$

$a$ and $a+1$ cannot share any divisors so there exists an $n$ where $an=(q-1)$ and  $(p-1)=n(a+1)$. 

Solve to $p$ and $q$ to get $q = na+1$ and $p=na+n+1=q+n$. with $b=ap=na^2+na+a$, the pair becomes: $(a, na^2+na+a)$ for $a$ and $n$ any integer. These pairs contain all possible pairs.