2007 number 5 - divisibility

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2007 number 5 - divisibility

Unread post by Katy729 » Sat Jun 24, 2017 3:31 pm

Let $a$ and $b$ be positive integers. Show that if $4ab-1$ divides $(4a^2-1)^2$, then $a=b$.

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Atonu Roy Chowdhury
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Re: 2007 number 5 - divisibility

Unread post by Atonu Roy Chowdhury » Fri Mar 30, 2018 10:22 pm

$4ab-1|(4a^2-1)^2 \Rightarrow 4ab-1|(a-b)^2$. Assume contradiction that $a \neq b$. wlog, $a>b$
$\frac{(a-b)^2}{4ab-1}=k \Rightarrow (a-b)^2=4abk-k$. Let $a$ and $b$ be such a solution such that $a+b$ is minimal.
Consider the quadratic equation
$\Rightarrow x^2 - x(2b+4bk)+(b^2+k)=0$. Obviously $a$ is a solution to this equation.
The other solution is $\frac{b^2+k}{a}$.
As $a+b$ is minimal, $\frac{b^2+k}{a} \geq a \Rightarrow k \geq a^2 - b^2$
$\frac{(a-b)^2}{4ab-1}=k \geq a^2 - b^2 \Rightarrow (a-b)^2 > (a+b)(a-b)$
implies $a-b > a+b$ which is obviously a contradiction.
So, $a=b$

EDIT: I just realized that $4$ is a dummy number here. Which means for all $n \in \mathbb{N}$ and $n>1$ if $nab-1|(na^2-1)^2$, then $a=b$
This was freedom. Losing all hope was freedom.

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