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Perfect Square ratio

Posted: Sun Nov 09, 2014 4:06 pm
by Fm Jakaria
Determine all ordered pair of positive integers $(x,y)$ so that
$\frac{x^2+2y^2}{2x^2+y^2}$ is the square of an integer.

Re: Perfect Square ratio

Posted: Mon Nov 10, 2014 3:32 am
by Nirjhor
If $x\neq y$ then \[\mathbb{N}\ni\dfrac{x^2+2y^2}{2x^2+y^2}=1+\dfrac{y^2-x^2}{2x^2+y^2}~\Rightarrow ~ 2x^2+y^2\mid y^2-x^2\] \[\Rightarrow ~ 2x^2+y^2\le y^2-x^2 ~\Rightarrow ~ 3x^2\le 0 ~\Rightarrow ~ x=0\not\in\mathbb{N} \] hence we must have $x=y$, thus all pairs $\mathbb{N}^2\ni (x,y)=(n,n)$ works.

Re: Perfect Square ratio

Posted: Mon Nov 10, 2014 7:06 pm
by mutasimmim
The reasoning falls apart if $ x $ is greater than $ y $.

Re: Perfect Square ratio

Posted: Mon Nov 10, 2014 7:46 pm
by Nirjhor
mutasimmim wrote:The reasoning falls apart if $ x $ is greater than $ y $
which is impossible.

Re: Perfect Square ratio

Posted: Mon Nov 10, 2014 7:50 pm
by Nirjhor
And that can be shown in two different ways.

Re: Perfect Square ratio

Posted: Thu Nov 13, 2014 2:38 pm
by SANZEED
To avoid the confusion, we may write that $x^{2}+2y^{2}\leq |y^{2}-x^{2}|$. Square it, simplify it, then we can get $3x^{2}(x^{2}+2y^{2})\leq 0$.