## 2007 number 5 - divisibility

Discussion on International Mathematical Olympiad (IMO)
Katy729
Posts: 47
Joined: Sat May 06, 2017 2:30 am

### 2007 number 5 - divisibility

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

Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm
Location: Chittagong, Bangladesh

### Re: 2007 number 5 - divisibility

$4ab-1|(4a^2-1)^2 \Rightarrow 4ab-1|(a-b)^2$. 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
$$(x-b)^2=4xbk-k$$
$\Rightarrow x^2 - x(2b+4bk)+(b^2+k)=0$. 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.