Imo 2007-5

[Masum]

Prove that if $4ab-1$ divides $(4a^2-1)^2$,then $a=b$

[tanvirab]

Ah, this problem kept me occupied for more than three hours in Vietnam and still did not surrender.

[Masum]

Hint: $4ab-1|(4a^2-1)^2$,then $4ab-1|(4a^2-1)^2-(4ab-1)^2=4a(a-b)(4a^2+4ab-2),4ab-1|(a-b)(4a^2-1)$(why?) and then $4ab-1|(a-b)^2$

[tarek like math]

$(4a^2-1)=A$ $(4ab-1)=B$ now $B$ factor of $A^2$ so $A,B$ have one or more common factors. suppose $d$ factor of $A$ and $B$. then $d/A,B=d$ or, $d/B-A=d$ so, $d/4ab-4a^2$. $d$ not equal to $4$,bcz $A=-1(mod)4$ so $d$ factor of $a$. suppose $a=nd$. now $A=4(nd)^2-1$ $B=4ndb-1$. now $A=-1(mod)d$ $B=-1(mod)d$. contradiction!
so $d$ not factor of $A,B$. but if $B$ factor of $A^2$ then $A,B$ have common factor like $d$ then $(B-A)$ divisible by $d$. so only way is $d$ not factor of $(B-A)$ this happen if $(B-A)=0$ then $4ab=4a^2$ then $a=b$.

[tanvirab]

Your first line seems incorrect.

[tarek like math]

if $B$ divide $A^2$ then $A,B$ must have common divisor. let $A=P1^(x1).P2^(x2)..Pn^(xn)$ $A^2=P1^(2x1)P2^(2x2)...Pn^(2xn)$ if $A^2=B.m$ then sqr root of $B.m=P1^(x1).P2^(x2)...Pn^(xn)$. since $B.m$ integer and they sharing same prime factors of $A$ then it seems $A,B,m$ have some common divisors. remember an integer can show as prime factor only in one way.

[tanvirab]

That's not the problem. How did you get $(4a^2−1)=A (4ab−1)$?

[Masum]

$(4a^2-1)=A$ $(4ab-1)=B$ now $B$ factor of $A^2$ so $A,B$ have one or more common factors. suppose $d$ factor of $A$ and $B$. then $d/A,B=d$ or, $d/B-A=d$ so, $d/4ab-4a^2$. $d$ not equal to $4$,bcz $A=-1(mod)4$ so $d$ factor of $a$. suppose $a=nd$. now $A=4(nd)^2-1$ $B=4ndb-1$. now $A=-1(mod)d$ $B=-1(mod)d$. contradiction!
so $d$ not factor of $A,B$. but if $B$ factor of $A^2$ then $A,B$ have common factor like $d$ then $(B-A)$ divisible by $d$. so only way is $d$ not factor of $(B-A)$ this happen if $(B-A)=0$ then $4ab=4a^2$ then $a=b$.

You already assumed \[4ab-1|4a^2-1\]
But if so, then this is simply trivial.
\[4ab-1|4a^2-1\]
\[\Longrightarrow 4ab-1|4a^2-4ab=4a(a-b)\]
But $\gcd(4ab-1,4a)=1$.
So \[4ab-1|a-b\]
But $a-b<4ab-1$
So, $a-b=0,a=b$

[tarek like math]

@Masum, i don't assume $(4ab-1)$ divisor of $4a^2-1$. i say if $4ab-1$ divisor of $(4a^2-1)^2$ then $4ab-1$ and $4a^2-1$ have common divisor suppose $d$. then $d$ divisor of $(4ab-1)-(4a^2-1)$. then it makes a contradiction. which only can ignore if we accept $a=b$. u also can see my reply of Tanvirab.
@Tanvirab, it is actually suppose $(4a^2-1)=A$, $(4ab-1)=B$

[Masum]

@Masum, i don't assume $(4ab-1)$ divisor of $4a^2-1$. i say if $4ab-1$ divisor of $(4a^2-1)^2$ then $4ab-1$ and $4a^2-1$ have common divisor suppose $d$. then $d$ divisor of $(4ab-1)-(4a^2-1)$. then it makes a contradiction. which only can ignore if we $\boxed{accept}$ $a=b$. u also can see my reply of Tanvirab.

What is that? I didn't understand.
There is no contradiction. It means that if $d$ is a common divisor of $4a^2-1$ and $4ab-1$, then it is a common divisor of $a-b$ too.