International Mathematical Olympiad

Proof Call $(a,b)$ a “bad pair” if it satisfies $4ab-1\mid (4a^2-1{)}^2$ while $a\ne b$. We use the method of infinite descent to prove there is no such “bad pair”.

Property 1 If $(a,b)$ is a “bad pair” and $a<b$, there exists an integer $c$ ($c<a$) such that $(a,c)$ is also a “bad pair”. In fact, let $r=\frac{(4a^2-1{)}^2}{4ab-1}$, then $$r=-r\cdot (-1)=-(4a^2-1{)}^2=-1(\mathrm{mod}4a).$$ Therefore there exists an integer $c$ such that $r=4ac-1$. Since $a<b$, we have $$4ac-1=\frac{(4a^2-1{)}^2}{4ab-1}<4a^2-1.$$ So $c<a$ and $4ac-1\mid (4a^2-1{)}^2$. Thus $(a,c)$ is a “bad pair” too.

Property 2 If $(a,b)$ is a “bad pair”, so is $(b,a)$. In fact, by $$1=1^2\equiv (4ab{)}^2(\mathrm{mod}4ab-1),$$ we get $$\begin{gathered} (4b^2-1{)}^2\equiv (4b^2-(4ab{)}^2{)}^2=16b^4(4a^2-1{)}^2 \\ \equiv 0(\text{mod }4ab-1). \end{gathered}$$ Thus $4ab-1\mid (4b^2-1{)}^2$.

In the following we will show that such “bad pair” does not exist. We shall prove by contradiction. Suppose there is at least one “bad pair”, we choose such pair for which $2a+b$ is minimum. If $a<b$, by property 1, there is a “bad pair” $(a,c)$ which satisfies $c<b$, and $2a+c<2a+b$, a contradiction. If $b<a$, by property 2, $(b,a)$ is also a “bad pair”, which leads to $2b+a<2a+b$, a contradiction. Hence such “bad pair” does not exist. Therefore $a=b$.