Does there exist a natural number pair $(a,b)$ such that $a^2b+a+5$ divide $ab^2+a+b$?

Since $a^2b+a+5$ divides $ab^2+a+b$, we can conclude that it also divides: $$a(ab^2+a+b)-b(a^2b+a+5)=a^2-5b$$ This means that we must have $a^2-5b=0$ or we must have $|a^2-5b| geqslant a^2b+a+5$. In the first case, substituting $a=5k$ and $b=5k^2$ gives: $$(125k^4+5k+5) mid (125k^5+5k^2+5k)$$ which is obviously true. Next, in the second case, we either have: $$|a^2| geqslant |a^2-5b| implies a^2 geqslant a^2b+a+5$$ which is impossible, or we have: $$5b geqslant |a^2-5b| geqslant a^2b+a+5 implies a leqslant 2$$ If $a=1$, then $b+6$ divides $b^2+b+1$. Performing the same deleting technique gives $b+6 mid b+37 implies b+6 mid 31$ which gives $b=25$.

If $a=2$, then $4b+7$ divides $2b^2+b+2$, and the trick gives $(2,11)$. Can you see that?

Thus, the only solutions are: $$(a,b)=(5k,5k^2),(1,25),(2,11)$$ for positive integers $k$.

Motivation : We must try to delete the first term in the number divisible by $ab^2+a+b$ so that we can perform bounding.