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ganeshie8

  • one year ago

show that \[\large \dfrac{3^n-1}{2^n-1}\] is never an integer for \(n\gt 1\) and \(n\in\mathbb{Z}\)

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  1. anonymous
    • one year ago
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    can u make it clear pls

  2. ganeshie8
    • one year ago
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    hey please check now @chethus

  3. anonymous
    • one year ago
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    still not clear

  4. ganeshie8
    • one year ago
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    basically the problem is about showing that expression always evaluates to some non integer

  5. anonymous
    • one year ago
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    i should know the statement rite?

  6. anonymous
    • one year ago
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    *

  7. dan815
    • one year ago
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    even/odd?

  8. dan815
    • one year ago
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    thats a start

  9. ganeshie8
    • one year ago
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    looks like a good start! i know \(3\mid (2^n-1)\) when \(n\) is even

  10. dan815
    • one year ago
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    |dw:1435130862127:dw|

  11. anonymous
    • one year ago
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    It will be a rational number if \[2^n-1\neq0\] as both numerator and denominator will be integers clearly \[2^n \neq 1\]\[n \log2 \neq \log1\]\[n \log 2 \neq 0\]\[n \neq 0\] which is satisfied

  12. dan815
    • one year ago
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    |dw:1435130938137:dw|

  13. dan815
    • one year ago
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    ok i see why 3 divides it now

  14. ganeshie8
    • one year ago
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    woah you always find a way to use binomial theorem !

  15. ikram002p
    • one year ago
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    when n is even :- 3|2^n-1 thus 2^n-1=3k bur 3 do not divide 3^n-1 3^n-1 cant be of the form 3f

  16. dan815
    • one year ago
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    lols its all i know

  17. anonymous
    • one year ago
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    ganeshie can u check my post?see above

  18. ganeshie8
    • one year ago
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    im pretty sure you can derive below identity with binomial thm too \[4^n-1^n = (4-1)(4^{n-1}+4^{n-2}+\cdots+1) = 3(stuff)\]

  19. anonymous
    • one year ago
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    no wait it seems wrong XD

  20. ganeshie8
    • one year ago
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    @Nishant_Garg yes the given expression is always a rational number for \(n\gt 1\) but that doesn't prove it can never be an integer right

  21. dan815
    • one year ago
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    i think this means we solved it for even atleast

  22. ikram002p
    • one year ago
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    huh duh

  23. dan815
    • one year ago
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    we dont need u here U OS HATER

  24. anonymous
    • one year ago
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    hmmmm

  25. ganeshie8
    • one year ago
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    haha when ppl say i hate myself, they don't literally mean it

  26. ikram002p
    • one year ago
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    in general k^n-1 do not divide g^n-1 for gcd(k,g)=1 both primes but that need primitive thingy

  27. ikram002p
    • one year ago
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    no no no dan i dont hate os :O

  28. dan815
    • one year ago
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    |dw:1435131204394:dw|

  29. ikram002p
    • one year ago
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    ok i have other idea

  30. ganeshie8
    • one year ago
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    so are you saying \[\large \dfrac{g^n-1}{k^n-1}\] is never an integer when \(\gcd(g,k)=1\) ?

  31. ganeshie8
    • one year ago
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    @ikram002p

  32. ikram002p
    • one year ago
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    yes

  33. ikram002p
    • one year ago
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    well no -.-

  34. dan815
    • one year ago
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    |dw:1435131317714:dw|

  35. ikram002p
    • one year ago
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    only when k is 2 (sorry not focused )

  36. ganeshie8
    • one year ago
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    okay odd case looks a bit more involved

  37. dan815
    • one year ago
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    the bottom is 2 mod 3

  38. dan815
    • one year ago
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    |dw:1435131616477:dw|

  39. ganeshie8
    • one year ago
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    top is 2 mod 3 bottom is 1 mod 3

  40. ikram002p
    • one year ago
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    cool

  41. ganeshie8
    • one year ago
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    i didn't get how that helps

  42. dan815
    • one year ago
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    oh nvm thats just silly

  43. dan815
    • one year ago
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    |dw:1435131948602:dw|

  44. dan815
    • one year ago
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    |dw:1435132054057:dw|

  45. dan815
    • one year ago
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    |dw:1435132120162:dw|

  46. dan815
    • one year ago
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    |dw:1435132232942:dw|

  47. dan815
    • one year ago
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    does any of this help

  48. ikram002p
    • one year ago
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    can we try mod 5 -.-

  49. dan815
    • one year ago
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    or 7

  50. dan815
    • one year ago
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    can we do induction maybe

  51. UsukiDoll
    • one year ago
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    @chethus this is a proof problem (pure math) which means we have to use different definitions or theorems to prove or disprove a statement. At first, it's hard to grasp because pure math is way different than computation math which is easily found in elementary, middle, or high schools

  52. dan815
    • one year ago
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    what does fermats little theorem say again

  53. UsukiDoll
    • one year ago
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    If p is a prime number then for any integer a , the number \[a^p-a\] is an integer multiple of p. IN the notation of modular arithmetic. this is expressed as \[a^p \equiv a (\mod p) \]

  54. ikram002p
    • one year ago
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    i give up, have no clue to start with.

  55. UsukiDoll
    • one year ago
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    If a is not divisible by p, Fermat's Little Theorem is equivalent to the statement that \[a^{p-1} -1 \] is an integer multiple of p or in symbols \[a^{p-1} \equiv 1 (\mod p)\]

  56. ikram002p
    • one year ago
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    bbye every one :)

  57. UsukiDoll
    • one year ago
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    \[\large \dfrac{3^n-1}{2^n-1} \] looks similar to the a not divisible by p . I feel like letting n =p-1 but that's just silly

  58. dan815
    • one year ago
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    did u find a solution to this yetq

  59. UsukiDoll
    • one year ago
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    no. T_T

  60. ganeshie8
    • one year ago
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    me neither, but i feel there exists a simple way for the odd case too

  61. UsukiDoll
    • one year ago
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    who came up with this T_T

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