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October 8th, 2017, 06:59 AM   #1
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Stuck in problem with level surfaces.

Given is the function f: R^2 -> R, with f(x,y)=x^2+y^2-6xy+8y
The level surface f(x,y)=1 contains infinitely much points (x,y) where x and y
are integer.
How can I prove this?
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October 8th, 2017, 01:41 PM   #2
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I'd start by rotating the system via

$\begin{pmatrix}x' \\ y'\end{pmatrix} = \dfrac{1}{\sqrt{2}}\begin{pmatrix}1 &-1\\1 &1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} $

That gets you a standard quadratic form you can play with.
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October 8th, 2017, 11:46 PM   #3
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Quote:
Originally Posted by romsek View Post
I'd start by rotating the system via

$\begin{pmatrix}x' \\ y'\end{pmatrix} = \dfrac{1}{\sqrt{2}}\begin{pmatrix}1 &-1\\1 &1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} $

That gets you a standard quadratic form you can play with.
Thank you for your reaction, but would you like to help me a little bit more?
I don't understand why rotating the system helps me prove this.
I understand rotation with a matrix, but I don't know how I rotate my function with 2 variables.

If I get a standard quadratic function, then I have an idea how I can prove this.
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October 9th, 2017, 03:20 AM   #4
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October 9th, 2017, 03:49 PM   #5
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Math Focus: Dynamical systems, analytic function theory, numerics
Suppose you have a quadratic equation of the form
\[
(x-a)^2 + (y-b)^2 = c.
\]
Can you see some relatively simple restrictions on $a,b,c$ which require this equation to have infinitely many integral solutions?
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October 10th, 2017, 09:04 AM   #6
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Quote:
Originally Posted by SDK View Post
Suppose you have a quadratic equation of the form
\[
(x-a)^2 + (y-b)^2 = c.
\]
Can you see some relatively simple restrictions on $a,b,c$ which require this equation to have infinitely many integral solutions?
I am trying to prove it with the formula of Pell. Thank you
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October 10th, 2017, 11:04 AM   #7
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I have this already
(x-3y)^2-2*(2y-1)^2=-1

Call m=x-3y and n=2y-1.
Then I recognize the Pell theorem: n^2-2p^2=-1.

This equation has a integer solution for n=p=1.
But how can I find the other solutions with the Pell theorem of n and p? (and x and y?)
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October 10th, 2017, 12:31 PM   #8
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Quote:
Originally Posted by Elize88 View Post
I have this already
(x-3y)^2-2*(2y-1)^2=-1

Call m=x-3y and n=2y-1.
Then I recognize the Pell theorem: n^2-2p^2=-1.

This equation has a integer solution for n=p=1.
But how can I find the other solutions with the Pell theorem of n and p? (and x and y?)
look at this
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