# 11^n = a Fibonacci Identity Problem

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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) )

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• 4 weeks later...

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

Theorem

For $A,B,x,y,m \in Z | (A,B = Constants)$

Let $N = F_{x,y} = x^{2} + Bxy -y^{2}$ where gcd(x,y) = 1

Then there exists a coprime pair $(x_{1},y_{1})$ such that

$F_{x_{1},y_{1}} = N^{2^{m}}$

I have a recursive formula that gives $(x_{1},y_{1})$

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

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

e.g.

$2^{2} - 1^{2} = 3$

$5^{2} - 4^{2} = 3^{2}$

$41^{2}-40^{2}=3^{4}$

$3281^{2}-3280^{2}=3^{8}$

$3^{2}-2^{2}=5$

$13^{2}-12^{2}=5^{2}$

$313^{2}-312^{2}=5^{4}$

$195313^{2}-195312^{2}=5^{8}$

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