Beautifull divisibility

For discussing Olympiad Level Number Theory problems
Katy729
Posts: 40
Joined: Sat May 06, 2017 2:30 am

Beautifull divisibility

Determine all pairs $(a,b)$ of positive integers such that $a^{2}b+a+b$ is divisible by $ab^{2}+b+7$

Abdullah Al Tanzim
Posts: 9
Joined: Tue Apr 11, 2017 12:03 am

Re: Beautifull divisibility

$ab^2+b+7 |a^2b+a+b$
$Case\ 1: when \ a<b$
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 $(7k^2,7k),(7,0)$
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