<s>[INST] Does there exist a natural number pair $(a,b)$ such that $a^2b+a+5$ divide $ab^2+a+b$? I tried that if I can show $\gcd(a^2b+a+5,ab^2+a+b) = 1$, then there is no solution. Am I true?
Any hints /idea?
I can't factorize them out such that each is multiple of each other.
 [/INST] 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.
</s>