Imo 1988,6

[Masum]

My most favourite problem from IMO (though historical).
Prove that if $a,b,\frac{a^2+b^2}{ab+1}=k\in\mathbb N$, then $k$ is a perfect square.

[Tahmid Hasan]

i have seen the solution to a similar problem.
$\frac{x^2+y^2+1}{xy}=k$ and $k$ is a natural number,then the only value of $k$ is $3$.
hint:let's flip some roots.

[mahathir]

I solved it by division.At first,if $k,a,b$ G.C.D is $1$,then, $a=b=k=1$.Again,if they are divisible by any other number,then,if their G.C.D is $d$,then,$k$ will be a number like $d^2*t$.Then,$a=dy$for some positive integer $d$ and $y$ and $b=dz$,for some positive integer $z$.now,we see here,that since G.C.D of $t,y,z$ is $1$,so,$t=z=y=1$.
Thus,$k$ will have to be a perfect square.
(i did not not show the equations,because they are pretty straightforward)

[Masum]

I solved it by division.At first,if $k,a,b$ G.C.D is $1$,then, $a=b=k=1$

Really? Show with proof.

(i did not not show the equations,because they are pretty straightforward)

They aren't straight forward at all.

Also, if $\gcd(t,x,y,z)=1$, who says that $t$ must be $1$?

[mahathir]

The equation can be written as $b^2-k=kab-a^2$ and
$a^2-k=kab-b^2$
from this,we understand if $(a,b)=1$ that is they r co-prime,then,we get,
$b^2$ is congruent to $kmoda$
$a^2$ is congruent to $kmodb$
from this,after some steps,we get,$k$ is congruent to $ab mod 1$
so,$k=ab$.but as they are co-prime,so,$b=1$ and $a=1$.so we get,$a=b=k=1$.

[Masum]

from this,after some steps,we get,$k$ is congruent to $ab mod 1$
so,$k=ab$

From whom did you learn this?

[*Mahi*]

Vieta jumping!
http://en.wikipedia.org/wiki/Vieta_jumping