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Subject: "11^8, a Fibonacci problem!"     Previous Topic | Next Topic
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Kenneth Ramsey
guest
Feb-19-07, 08:59 AM (EST)
 
"11^8, a Fibonacci problem!"
 
   Problem: Use Fibonacci like series (but not the same starting numbers) to prove that there are an infinite number of solutions to 5a^2 + 5ab + b^2 = 11^8 with a and b coprime. In I have a proof that you can solve for any prime, ending in 1 or 9, multiplied to the 8th power.


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Kenneth Ramsey
guest
Feb-23-07, 06:37 AM (EST)
 
1. "RE: 11^8, a Fibonacci problem!"
In response to message #0
 
   I have a proof that P = 5a^2 + 5ab + b^2 has solutions where a and b are coprime if and only if P is not divisible by any prime ending in 2,3,5 or 7. My proof provides an algorithim for finding a and b using determinants involving prior solutions sets for the prime factors of P that end in 1 or 9. The proof can be understood by reference to elementary algebra and the law of quadratic reciprocity.

For instance by knowing that a = 1, b = 1 is a solution set for P = 11, it is possible to calculate that a = 3161, b = 7159 is a solution set for 11^8 power in just 3 iterations and to further find a solution for P = 11^9 in at most 5 iterations. It takes just one iteration to go from a solution set for going from a solution set for p^n to p^2n but may take 2 iterations for going from solution sets for p^A and p^B to p^(A+B) since it is possible to unknowingly pick a course that gives p^(A+B) where a and b are not coprime, in that case one can revert to a course that will assure a coprime solution set.

Due to a lack of response I am loosing interest in posting on these forums. Please see my further posts at my own forum on Fibonacci and Triangular numbers at https://groups.yahoo.com/group/Triangular_and_Fibonacci_Numbers Be careful to take into account that this link can be inavertently parsed by the forum software.


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