Proving an Infinite Number of Integer Points on a Level Surface

[Elize88]

Given is the function f: R² -> R, with f(x,y)=x²+y²-6xy+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?

I see that it is true with some examples, but how can I prove this.
Do I need to use the gradient? Or tangent planes? Or linear algebra?

 

[Opalg]

Given is the function f: R² -> R, with f(x,y)=x²+y²-6xy+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?

I see that it is true with some examples, but how can I prove this.
Do I need to use the gradient? Or tangent planes? Or linear algebra?

Hi Elize, and welcome to MHB!

You want to find integer solutions to the equation $x^2 - 6xy +y^2 + 8y = 1$. I would start by completing the square and writing the equation as $(x - 3y)^2 = 8y^2 - 8y + 1$. If you can find an integer value of $y$ making the right-hand side of that equation a perfect square, say $8y^2 - 8y + 1 = n^2$, then $(x - 3y)^2 = n^2$. So $x = 3y\pm n$, giving you two integer values of $x$ to solve your equation.

But $8y^2 - 8y + 1 = 2(2y-1)^2 - 1$. So we want to find integer solutions of the equation $n^2 = 2(2y-1)^2 - 1$. In other words, we are looking for squares that are equal to twice a square minus $1$. That should get you thinking about http://mathhelpboards.com/math-notes-49/pell-sequence-2905.html.