Perfect Square ratio

For discussing Olympiad Level Number Theory problems
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Fm Jakaria
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Perfect Square ratio

Unread post by Fm Jakaria » Sun Nov 09, 2014 4:06 pm

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.
You cannot say if I fail to recite-
the umpteenth digit of PI,
Whether I'll live - or
whether I may, drown in tub and die.

Nirjhor
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Re: Perfect Square ratio

Unread post by Nirjhor » Mon Nov 10, 2014 3:32 am

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.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


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mutasimmim
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Re: Perfect Square ratio

Unread post by mutasimmim » Mon Nov 10, 2014 7:06 pm

The reasoning falls apart if $ x $ is greater than $ y $.

Nirjhor
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Location:Varies.

Re: Perfect Square ratio

Unread post by Nirjhor » Mon Nov 10, 2014 7:46 pm

mutasimmim wrote:The reasoning falls apart if $ x $ is greater than $ y $
which is impossible.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

Nirjhor
Posts:136
Joined:Thu Aug 29, 2013 11:21 pm
Location:Varies.

Re: Perfect Square ratio

Unread post by Nirjhor » Mon Nov 10, 2014 7:50 pm

And that can be shown in two different ways.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

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SANZEED
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Location:Mymensingh, Bangladesh

Re: Perfect Square ratio

Unread post by SANZEED » Thu Nov 13, 2014 2:38 pm

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 $3x^{2}(x^{2}+2y^{2})\leq 0$.
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