1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Natural number question

  1. Jan 14, 2017 #1
    1. The problem statement, all variables and given/known data
    Prove that there doesnt exist natural numbers x and y such that the statement holds true.
    y^5 + 1 = (x^7-1)/(x-1)


    3. The attempt at a solution
    I was able to simplify the term down to
    y^5 / x = x^5 + (x^5-1)/(x-1)
    Not sure what to do with it
     
  2. jcsd
  3. Jan 14, 2017 #2

    Math_QED

    User Avatar
    Homework Helper

    I'm not sure but maybe you can use this:

    Note that ##\frac{x^7 - 1}{x - 1} = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1##.
     
  4. Jan 14, 2017 #3
    Yes in that form if x is not equal to 3k -1 it is true but if it is its not coming true
     
  5. Jan 14, 2017 #4
    y5 = x(x+1)(x4+x2+1)
    gcd(x,x+1) = gcd(x,x4+x2+1 )=1
    however
    gcd(x+1,x4+x2+1)=gcd(x+1,3)
     
  6. Jan 14, 2017 #5
    How should I proceed?
     
  7. Jan 17, 2017 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Consider a prime factor p of x. Suppose it divides x exactly k times. What can you deduce about y in relation to p?
     
    Last edited: Jan 17, 2017
  8. Jan 17, 2017 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    @Faiq , have you lost interest in this? My method works.
     
  9. Jan 18, 2017 #8
    Don't know about faiq but I am interested in this. All I can infer from your hint is that y has also all the prime divisors of x at least once. After that what follows?
     
  10. Jan 18, 2017 #9

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Can you be more exact?
     
  11. Jan 18, 2017 #10
    Nope, I don't think I can prove that y has all the prime divisors of x exactly the same times that is essensially that y is a multiple of x? is this what you trying to tell me?
     
  12. Jan 18, 2017 #11

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Almost.
    If p divides x exactly k times, how often does it divide y5? What does that tell you about k?
     
  13. Jan 18, 2017 #12
    It divides ##y^5## at least k times but I got no clue what does this tells me about ##k##???
     
  14. Jan 18, 2017 #13

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Can it divide y5 more than k times?
     
  15. Jan 19, 2017 #14
    I guess not if I take into account some of the posts of @Faiq about the gcd..

    Sorry but anyway I don't see where you trying to lead me, but I just wanna know what theorems/lemmas of number theory you are using. Do you use only the fundamental theorem of number theory along perhaps with some of its immediate corollaries (would be good to mention which ones you use though).
     
  16. Jan 19, 2017 #15

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    There are two factors each side: (y5)(x-1)=(x)(x6-1). If p divides x k times but y5 more than k times, p needs to divide the other factor on the right. Is that possible?
    So what can we say about the relationship between x and y5? What about the other factors of y5?

    I am not using any theorems or lemmas, just simple logic.
     
  17. Jan 19, 2017 #16
    Seems to me you are using some lemmas besides the fundamental theorem of number theory but for you must be so obvious and simple facts that you call them simple logic. However for me they are not so obvious. For example you seem to use a little lemma that "if a prime p divides a product ab then p divides a or p divides b" which is a corollary from the fundamental theorem.

    (I wonder why it seems so obvious to you that p cannot divide ##x^6-1##)
    ok anyway seems to me you implying that ##y^5=cx## where c contains the other prime factors of y^5. But still cant see where do we go from that. c seems complete mystery to me I cant make anything about it. We want to prove that y cannot be integer if x is integer right?
     
  18. Jan 19, 2017 #17

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The multiples of p are at intervals of p, so two consecutive numbers cannot have a factor in common.

    Since c and x are coprime, and their product is a fifth power, what does that tell you about c and x individually?
     
  19. Jan 20, 2017 #18
    That's another little lemma that you classify as simple logic I guess (hehe), if x and x+1 have a common factor then their difference which is 1 must be a multiple of their common factor, ok got it.
    Here is the critical piece of the proof for me, I couldn't even imagine or notice that c and x are coprime...If I understand correctly since their product is a fifth power it means that they aren't coprime afterall which is a contradiction correct?
     
  20. Jan 20, 2017 #19

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Right.
    Wrong.
     
  21. Jan 20, 2017 #20
    Hmmm ok it means that they are both a fifth power. But if x is a fifth power then c (which is ##(x+1)(x^4+x^2+1)##) cannot be a fifth power?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Natural number question
  1. Nature of this number (Replies: 13)

  2. Natural number (Replies: 35)

Loading...