$ab^2+b+7 |a^2b+a+b$
$Case\ 1: when \ a<b$
from divisibility inequality we know that,
$a^2b+a+b \geq ab^2+b+7$
or, $ab(a-b) \geq 7-a$
or, $a-7 \geq ab(b-a)$
it is clear that there is no solution for $a,b\ \epsilon\ N\ if\ a<b$
$ Case\ 2: when\ a=b$
$ ab(a-b)=7-a $
or, $ a(ab-b^2+1)=7 $
from that $ (7,7)$ and $ (7,0) $ are two solutions.
$Case\ 3 : when\ a>b$
$ b(ab+1)+7 | a(ab+1)+b $
$If\ b|a$ then $7|b$ and it is also true that if $a=bk$ then $b=7k$. So ,$ a=7k^2$ for $K\ \epsilon\ N $
so, the solutions are $(7k^2,7k),(7,0)$
Everybody is a genius.... But if you judge a fish by its ability to climb a tree, it will spend its whole life believing that it is stupid - Albert Einstein