Beautifull divisibility

For discussing Olympiad Level Number Theory problems
Facebook Twitter

Beautifull divisibility

Post Number:#1  Unread postby Katy729 » Sat Jul 01, 2017 3:44 pm

Determine all pairs $(a,b)$ of positive integers such that $a^{2}b+a+b$ is divisible by $ab^{2}+b+7$
Katy729
 
Posts: 31
Joined: Sat May 06, 2017 2:30 am

Re: Beautifull divisibility

Post Number:#2  Unread postby Abdullah Al Tanzim » Sun Jul 30, 2017 11:03 am

$ab^2+b+7 |a^2b+a+b$
$Case\ 1: when \ a<b$
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
User avatar
Abdullah Al Tanzim
 
Posts: 7
Joined: Tue Apr 11, 2017 12:03 am
Location: Dhaka, Bangladesh.


Share with your friends: Facebook Twitter

  • Similar topics
    Replies
    Views
    Author

Return to Number Theory

Who is online

Users browsing this forum: No registered users and 1 guest

cron