Ramsey2879 Posted February 22, 2007 Share Posted February 22, 2007 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) ) Link to comment Share on other sites More sharing options...
Ramsey2879 Posted March 17, 2007 Author Share Posted March 17, 2007 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] Link to comment Share on other sites More sharing options...
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