11^n = a Fibonacci Identity Problem

[Ramsey2879]

Actually my problem is to find my proof that there are an infinite number of solutions to the Diophantine equation 5a^2 + 5ab + b^2 = p^n where a,b are coprime and p is a prime ending in 1 or 9. There is a relationship between any three consecutive term of a Fibonacci type series and the form 5a^2 + 5ab + b^2 that is invariant with the index number of the first term. That is a key to my proof. I leave the proof for you to figure out, but will make suggestions if you reach a dead end and have no idea where to turn.

 

First off then, how does the form 5a^2 + 5ab + b^2 relate to three terms of a Fibonacci type series in an invariant manner? (By Fibonacci type, I mean F(n) = F(n-1) + F(n-2) )

[Ramsey2879]

My proof extends to still a more general case that includes Pythagorean Triples

Theorem

For [math]A,B,x,y,m \in Z | (A,B = Constants)[/math]

Let [math] N = F_{x,y} = x^{2} + Bxy -y^{2}[/math] where gcd(x,y) = 1

Then there exists a coprime pair [math](x_{1},y_{1})[/math] such that

[math]F_{x_{1},y_{1}} = N^{2^{m}}[/math]

 

I have a recursive formula that gives [math](x_{1},y_{1})[/math]

 

It even gives values if [math] N = F_{x,y} = Ax^{2} + Bxy -y^{2}[/math] e.g. [math](x_{1},y_{1})[/math] such that [math]F = N^{2}[/math]

 

If A is a perfect square and B = 0 then my formula relates to Pythagorean triples

e.g.

[math]2^{2} - 1^{2} = 3[/math]

[math]5^{2} - 4^{2} = 3^{2}[/math]

[math]41^{2}-40^{2}=3^{4}[/math]

[math]3281^{2}-3280^{2}=3^{8}[/math]

 

[math]3^{2}-2^{2}=5[/math]

[math]13^{2}-12^{2}=5^{2}[/math]

[math]313^{2}-312^{2}=5^{4}[/math]

[math]195313^{2}-195312^{2}=5^{8}[/math]