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BdMO Online Forum • View topic - Particular divisibility..

Particular divisibility..

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Particular divisibility..

Post Number:#1  Unread postby Katy729 » Sun May 21, 2017 2:46 am

Find all pairs $(a,b)$ of positive integers for which $a^2-b$ divide $b^2+a$, and $b^2-a$
Divides $a^2 + b$
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Re: Particular divisibility..

Post Number:#2  Unread postby Atonu Roy Chowdhury » Sat Jun 03, 2017 11:37 pm

Case 1: $a=b>1$

$a^2-a|a^2+a$ or $a^2-a|2a$. So, $a \le 3$. Checking the values, we get $a=2$ and $a=3$.

Case 2: $a \neq b$

WLOG $a>b$
$a^2-b|b^2+a \Rightarrow b^2+a \ge a^2-b \Rightarrow b(b+1) \ge a(a-1)$. As $a>b$, we get $b+1>a-1 \Rightarrow a-b<2 \Rightarrow a=b+1$.
$b^2-a|a^2+b \Rightarrow b^2-b-1|b^2+3b+1 \Rightarrow b^2-b-1|4b+2$ which is obviously true for $b=1$. Now consider $b>1$ and then $b^2-b-1 \le 4b+2$ which is true for $b \le 5$ . Checking, we get the only possible value of $b$ is $2$.

All the solutions are $(a,b)=(2,2),(3,3),(2,1),(3,2)$ and permutations.
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