MHB Proving an Infinite Number of Integer Points on a Level Surface

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The function f(x,y) = x² + y² - 6xy + 8y defines a level surface where f(x,y) = 1, which contains infinitely many integer points (x,y). To prove this, one can start by rewriting the equation as (x - 3y)² = 8y² - 8y + 1. Finding integer values for y that make the right side a perfect square leads to integer solutions for x. Specifically, the equation can be transformed to n² = 2(2y - 1)² - 1, prompting the search for integer solutions that satisfy this condition. This approach suggests a connection to Pell's equation, indicating a pathway to proving the infinite integer solutions.
Elize88
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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?
 
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Elize88 said:
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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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