How do I prove (or disprove) that the formula x^2-xy-y^2=(+-)1 has an infinite amount of integer solutions?

Question

How do I prove/disprove that the formula x^2-xy-y^2=(+-)1 has an infinite amount of integer solutions?
I've observed that the positive integer solutions for values x and y are all consecutive Fibonacci numbers. but have no way of proving it.

Answer 1

Suppose x^2-xy-y^2 = 1. Let X =x+y and Y=x and compute:



X^2-XY-Y^2 = (x+y)^2 - (x+y)x - x^2 = x^2 + 2xy + y^2 - x^2 - xy -x^2 = -x^2 + y^2 + xy = - (x^2-xy-y^2) = -1.



So from the solution (1,1) we get a new solution (2,1).

From the solution (5,3) we get a new solution (8,5), etc.



The new solutions have -1 rather than 1 on the right side of the equal sign.



Now, to finish, suppose x^2-xy-y^2 = -1, you need to try something similar for X and Y to see if you can get a new solution with 1 on the right side of the equal sign. Putting the two things together you will have found infinitely many solutions, and probably, given the right start, proved the connection to Fibonacci as well. I will let you work out the details.

Answer 2

f(n) = f(n-2) + f(n - 1)



-->



You are saying that x = f(n), y = f(n + 1) solves the equation?



x = 1, y = 1 --> 1 - 1 - 1 = -1, YES

x = 1, y = 2 --> 1 - 2 - 4 = -5, NO

x = 2, y = 1 --> 4 - 2 - 1 = 1, YES

x = 3, y = 2 --> 9 - 6 - 4 = -1, YES

x = 5, y = 3 --> 25 - 15 - 9 = 1, YES



...OK, so you can prove this by induction:



Assuming that this is true for:

x = f(n), y = f(n-1), I'll call this P(n)



We can show that it is ALSO true for:

x = f(n + 1), y = f(n), I'll call this P(n + 1)



-->

P(n) implies the following:



(f(n))² - f(n) * f(n - 1) - (f(n - 1))² = ±1



Now I want to see if P(n + 1) holds, I'll do this by evaluating the LHS:



(f(n + 1))² - f(n + 1) * f(n) - (f(n))²



But I can write f(n + 1) in terms of f(n) and f(n - 1)



-->



(f(n) + f(n - 1))² - (f(n) + f(n - 1)) * f(n) - (f(n))²

--> now multiply things out

f²(n) + 2f(n)f(n - 1) + f²(n - 1) - f²(n) - f(n - 1)) * f(n) - f²(n)

--> collecting like terms this leaves:

f²(n - 1) + f(n)f(n - 1) - f²(n)

--> this looks VERY similar to the expression in P(n)



multiply by negative (switch the 1st and last term)



- (f²(n) - f(n)f(n - 1) - f²(n - 1))



But I know from my assumption that P(n) held, that this inside part is ±1



So I have:

-(±1) = -/+1, therefor P(n + 1) holds



So I have the following theorem:



P(n) -> P(n + 1)



That is, if x = f(n), y = f(n - 1) is a solution, then x = f(n + 1), y = f(n) is ALSO a solution. This is where you get your sequential Fibonacci numbers.



The only thing left to do (but I already did it above) is to prove the base case. Well the base case is n = 1 -->

x = f(1), y = f(0) --> x = 1, y = 1 --> proved to be true above



So you can say that P(1) --> P(n), for ALL n >= 1



since P(1) proves P(2), P(2) proves P(3)...etc, this is standard proof by Weak Induction.



Therefore yes you are correct that consecutive Fibonacci numbers are integer solutions. But we have by NO means shown that these are the ONLY integer solutions, but there are certainly infinitely many of them.

Answer 3

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I have no idea, as I didn't go that far in school.



I just stopped by to see if you got any answers.



I have no idea if the answers are right or wrong, but I'm impressed anyway! :~)





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