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11^8 = a Fibonacci Problem

  1. Feb 19, 2007 #1
    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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  3. Feb 22, 2007 #2

    Gib Z

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    [tex]T_n=T_{n-1} + T_{n-2}[/tex]
    Factorise.
    [tex](\sqrt{5}a +\sqrt{5}b)^2 - 4b^2 = 11^8[/tex]. Perhaps that reminds you of something?
     
  4. Feb 22, 2007 #3

    Gib Z

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    O, quote: I have a proof that you can solve for any prime, ending in 1 or 9, multiplied to the 8th power.

    Well, 11^8 = 123456787654321, which satisfies your conditions.
     
  5. Feb 22, 2007 #4
    Actually the problem I meant 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)
     
  6. Feb 22, 2007 #5
    This is interesting, it factors to [tex]((\sqrt{5}a +\sqrt{5}b +2b) * ((\sqrt{5}a +\sqrt{5}b -2b) = 11^8[/tex]. However, these are not integers and I thought number theory dealt only with integers.
     
    Last edited: Feb 22, 2007
  7. Feb 22, 2007 #6

    matt grime

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    That's a strange thing to say. Not least because 'a' isn't an integer, it is an indeterminate expression, so your own first post fails your own criterion.

    Moving on from the stupidly pedantic, you're objecting to the use of sqrt(5), right? Well, if you think that sqrt(5) has no place in number theory then you're not going to get very far in the subject. Number theory might be set up to answer questions posed about Z, or more likely Q, but there is nothing that says it must stick only with those things, and indeed it doesn't.
     
  8. Feb 22, 2007 #7
    Ok but how would you get an solution in integers to my diophantine equation using that? If there is a way I would like to know. But since thare are Fibonacci identities that are known, it is not necessary to resort to the square root of 5 or the golden ratio to make my proof.
     
  9. Feb 22, 2007 #8

    matt grime

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    I did not say anything about your equation. However, I believe it is a common tactic in algebraic number theory that in order to find integral solutions to something one passes to a larger ring, such as the ring of integers in a number field. This is probably called something like Class Field Theory.
     
  10. Feb 22, 2007 #9
     
  11. Feb 22, 2007 #10

    matt grime

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    Nope. coprimality exists in the algebraic integers too: x and y a coprime if z divides x and z divides y implies z is a unit. Anyway, you were asking for integral solutions to some equation in a and b. At least that is how I read it. But a was an indeterminate. Not an integer. 2 is an integer. a isn't. Unless a=2. For example. If a is an integer in post 1 what integer is it?
     
  12. Feb 22, 2007 #11

    matt grime

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  13. Feb 22, 2007 #12
    But how do you get 5^a^2 + 5ab + b^2 from 5(a+b)^2 - 4b^2
     
  14. Feb 22, 2007 #13

    matt grime

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    Uh? Please point out where you think I claim to have got that at all.
     
  15. Feb 22, 2007 #14
    l a = 3161 b = 7159 , is one of infinitely many solutions in integers for 11 ^8. But any prime ending in 1 or 9 to any integer power can replace 11^8 and there would still be an infinite number of solutions in integers.
     
  16. Feb 22, 2007 #15
    You didn't post it but there was a post that used sqrt(5) in a manner that gets 5a^2 + 10ab - b^2 which is not my equation. Sorry

    I have to go now. Thanks for your help.
     
  17. Feb 22, 2007 #16
    Are all primes ending in 1 or 9 and their products members of a ring? My proof related to this topic is much broader than the original proposition since it covers all primes ending in 1 or 9 or their products. In short I have a proof that there are an infinite number of solutions to the Diophantine equation
    [tex]P = 5a^{2} + 5ab + b^{2}[/tex] where P is a prime ending in 1 or 9 or a product of such primes. Knowing the algorithm makes it easy to solve for any such product given initial solutions for the primes ending in 1 or 9. There is no need to do calculations greater than the square root of P. The proof is really simple in retrospect and can be understood by an algebra student familiar with the law of quadratic reciprocity. I doubt that it could be simplified by using Class Field Theory. Perhaps it may be a trival application of class theory but I don't know. I skipped college after taking the equivalant of two years of courses via night school (none in higher algebra or number theory) and am now retired. However, my interest in number theory is such that I would be willing to pay to audit a few courses such as Algebraic Number Theory once I take care of a few other things like building my dream home. The book I purchased is mostly Greek to me.
     
  18. Feb 23, 2007 #17

    matt grime

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    OF course they are: the integers, the rationals, the reals, complex numbers, algebraic integers, algebraic numbers.....
     
  19. Feb 23, 2007 #18

    Gib Z

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    perhaps the most obvious one, the primes >.<"

    EDIT: btw, im taking ring as a synonym for set, i haven't done rings yet but it looks like you were listing the sets primes were in.
     
  20. Feb 23, 2007 #19

    matt grime

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    A ring is not just a set.
     
  21. Feb 23, 2007 #20

    Gib Z

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    ahh then excuse my ignorance :)
     
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