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 October 8th, 2017, 05:59 AM #1 Newbie   Joined: Oct 2017 From: netherlands Posts: 4 Thanks: 0 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?
 October 8th, 2017, 12:41 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,590 Thanks: 1434 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. Thanks from topsquark
October 8th, 2017, 10:46 PM   #3
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 Originally Posted by romsek 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.

 October 9th, 2017, 02:20 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2274 How far did you get?
 October 9th, 2017, 02:49 PM #5 Senior Member   Joined: Sep 2016 From: USA Posts: 670 Thanks: 440 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? Thanks from Elize88
October 10th, 2017, 08:04 AM   #6
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 Originally Posted by SDK 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

 October 10th, 2017, 10:04 AM #7 Newbie   Joined: Oct 2017 From: netherlands Posts: 4 Thanks: 0 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?)
October 10th, 2017, 11:31 AM   #8
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 Originally Posted by Elize88 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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