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TrojanPoem

  • one year ago

If z, y were the roots of the equation : x^2 -6 x + 1 = 0 Prove that : (y^n + z^n) is a Positive integer which can be divided by 5 for any integer n

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  1. phi
    • one year ago
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    I would first solve for the roots. what are they?

  2. TrojanPoem
    • one year ago
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    3 (+ or -) 2 sqrt(2)

  3. ganeshie8
    • one year ago
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    \[x^2-6x+1=0\implies x^2=6x-1\] Since \(z,y\) are roots, we have \[y^2 = 6y-1\implies y^n=6y^{n-1}-y^{n-2}\tag{1}\] \[z^2=6z-1\implies z^n=6z^{n-1}-z^{n-2}\tag{2}\] Add them both and conclude by induction

  4. TrojanPoem
    • one year ago
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    Hmm, I have to solve it using Combinations

  5. TrojanPoem
    • one year ago
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    And permutations

  6. TrojanPoem
    • one year ago
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    Maybe this can help , The book has this solved example y = x^4 + 6 x^3 + 11 x^2 + 6 x prove that it can be divided by 24 and was solved like this

  7. TrojanPoem
    • one year ago
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    y = x^4 + x^3 + 5x^3 + 5x^2 + 6 x^2 + 6x y = x^3(x+1) + 5x^2(x+1) + 6x(x+1) y = x(x+1)(x^2 + 5x +6 ) y = x(x+1)(x+2(x+3) = (x+3)P4 (x+3)P4 / 4! = x+3C4 always positive integer so it can be divided by (24 = 4!)

  8. TrojanPoem
    • one year ago
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    Now, it asks to apply what I learnt in the question above.

  9. ganeshie8
    • one year ago
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    Thats pretty clever!

  10. ganeshie8
    • one year ago
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    Your book is using the fact that \(\binom{n}{r}\) is an integer. You could also view it like this : product of any \(n\) consecutive integers is divisible by \(n!\)

  11. TrojanPoem
    • one year ago
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    Yeah, I understood from the explanation which is too clear. The problem is to solve "try yourself" problem

  12. TrojanPoem
    • one year ago
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    Any ideas ?

  13. ganeshie8
    • one year ago
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    Not getting any ideas yet but im still trying to see if the induction method that i suggested before really works

  14. TrojanPoem
    • one year ago
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    Ok.

  15. dan815
    • one year ago
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    binomial theorem

  16. dan815
    • one year ago
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    since u are mentioned permutations and combinations

  17. TrojanPoem
    • one year ago
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    I think you are right.

  18. dan815
    • one year ago
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    look at (z+y)^n = ....

  19. TrojanPoem
    • one year ago
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    Y^n + z^n we will sum them , and maybe the result can help.

  20. dan815
    • one year ago
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    what will z^n+y^n be equal to from that binomial expansion

  21. dan815
    • one year ago
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    yeah play around with that, i think u can show its divisible by 5 fomr it

  22. TrojanPoem
    • one year ago
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    I failed , try this before.

  23. TrojanPoem
    • one year ago
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    I tried*

  24. TrojanPoem
    • one year ago
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    ((3+sqrt(2))^2 + (3-sqrt(2))^2) = 2 Singles

  25. phi
    • one year ago
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    ** (y^n + z^n) is a Positive integer which can be divided by 5 for any integer n *** I am having a problem with the divide by 5. if we do a simple numerical check with n=2 and y= 3+2 sqr(2), z= 3 - 2 sqr(2) I get y^2 + z^2 = 34

  26. ganeshie8
    • one year ago
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    yeah it doesn even work for n=1

  27. TrojanPoem
    • one year ago
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    You think 5 is a typo from the book ?

  28. TrojanPoem
    • one year ago
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    so It's 2 not 5 as 2 will work for any n

  29. TrojanPoem
    • one year ago
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    Any ideas ?

  30. dan815
    • one year ago
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    recheck the question

  31. phi
    • one year ago
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    I would assume the problem would be closer to the example you posted, and you start with factors that are consecutive. And then you play the same trick. But that is not the case here, so I have no clear idea of what is the purpose of this question is here.

  32. TrojanPoem
    • one year ago
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    It's written as I have copied it.

  33. ganeshie8
    • one year ago
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    if psble take a screenshot of as much as possible and attach

  34. TrojanPoem
    • one year ago
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    The second example was intended to show how the writer of the book thinks and what he wants in answering my problem

  35. TrojanPoem
    • one year ago
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    I can't transfer my photos to the PC or laptop. But tell me what you mean by as much as possible and I will fetch it.

  36. dan815
    • one year ago
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    maybe its x^2 -5 x + 1 = 0 then for every odd n z^n + y^n is divisible by 5

  37. TrojanPoem
    • one year ago
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    It's written as x^2 -6 x + 1 = 0 in the book. But fine , all I want is to know it's idea solve yours

  38. dan815
    • one year ago
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    k lets work on this new problem then x^2 -5 x + 1 = 0 then for every odd positive integer n z^n + y^n is divisible by 5

  39. dan815
    • one year ago
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    prove

  40. TrojanPoem
    • one year ago
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    Lol me ? IDK I am even stuck with 5 more in this book.

  41. dan815
    • one year ago
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    oh this is kidn of silly lol, the sqrt 29 in the roots will cancel out for ever odd integer

  42. dan815
    • one year ago
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    okay well we need a different question

  43. TrojanPoem
    • one year ago
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    Post here ?

  44. dan815
    • one year ago
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    yeah

  45. TrojanPoem
    • one year ago
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    Find the coefficient of x^n in the expansion of (1 + x + 2x^2 + 3x^3 + ..... + nx^n)^2

  46. dan815
    • one year ago
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    u can write out on liek a dot product and it will be obvious

  47. TrojanPoem
    • one year ago
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    Show me

  48. dan815
    • one year ago
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    uh wait no lemme see

  49. dan815
    • one year ago
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    |dw:1434805428709:dw|

  50. dan815
    • one year ago
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    so that series

  51. dan815
    • one year ago
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    |dw:1434805474806:dw|

  52. dan815
    • one year ago
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    |dw:1434805575985:dw|

  53. dan815
    • one year ago
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    n+(n-1)*(n+1)*(n)/2

  54. TrojanPoem
    • one year ago
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    WOW, seems nice although it's out of my curriculum.

  55. dan815
    • one year ago
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    no no u know this stuff, u know summation of 1+2+3.... +n that formula right

  56. dan815
    • one year ago
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    im just using that to get an expression for the summation

  57. TrojanPoem
    • one year ago
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    arithmetic ?

  58. dan815
    • one year ago
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    yeah

  59. TrojanPoem
    • one year ago
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    But I have never seen this E or epslion.

  60. dan815
    • one year ago
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    k ignore that look at the pattern that is coming up, here is a simple example look at a simple example (1+2x+3x^2)^2

  61. dan815
    • one year ago
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    |dw:1434805819564:dw|

  62. dan815
    • one year ago
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    u mulitply the first and last term then the 2nd and 2nd last term 3rd and 3rd last term and u add them all up to get the constant on the nth degree

  63. TrojanPoem
    • one year ago
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    Each time you are multiplying by x and increase 1 to the coefficeitn

  64. dan815
    • one year ago
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    maybe its easier to see if i write it out in polynomial form

  65. dan815
    • one year ago
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    |dw:1434806006840:dw|

  66. TrojanPoem
    • one year ago
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    At first I though of solving it as the sum of geometrical like this ( 1 + x+ x^2 + x^3 + x^n + x^2 + x^3 +...+ x^n + x^3 +... +x^n until x^n is alone ) but failed ( too long)

  67. dan815
    • one year ago
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    |dw:1434806044605:dw|

  68. dan815
    • one year ago
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    u do the same thing with your bigger question ull see a pattern and u can see what they add upto

  69. TrojanPoem
    • one year ago
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    Oh , clarify a bit this multiplication

  70. dan815
    • one year ago
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    because all other multiplications will give different degree term

  71. dan815
    • one year ago
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    hm

  72. dan815
    • one year ago
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    i dunno how else to say it

  73. TrojanPoem
    • one year ago
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    I got it to some extent , focusing with myself I will get full understanding of it.

  74. TrojanPoem
    • one year ago
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    Now there is more 4 questions , wanna try ?

  75. dan815
    • one year ago
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    okay try a similar problem, can u tell me waht the coefficient on the n-1 th degree term is

  76. dan815
    • one year ago
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    and sure

  77. TrojanPoem
    • one year ago
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    let me think

  78. dan815
    • one year ago
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    are you in highschool btw?

  79. TrojanPoem
    • one year ago
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    yeah, but - not to lie- I never took this until now.

  80. dan815
    • one year ago
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    is this AP algebra?

  81. asnaseer
    • one year ago
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    if z,y are indeed the roots of: \(x^2-6x+1=0\) then this implies that: \(z+y=6\implies z=6-y\) therefore: \(z^n+y^n=(6-y)^n+y^n=6^n-n6^{n-1}y+...-y^n+y^n\) Note that the last two terms cancel out leaving an expression that is always a multiple of 6

  82. asnaseer
    • one year ago
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    so it cannot be a multiple of 5 for all n - are you sure the question is stated correctly?

  83. TrojanPoem
    • one year ago
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    I wrote it as It was in the book but maybe typo of the book writer . But WOW you got it

  84. asnaseer
    • one year ago
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    looking at the other replies I believe @dan815 proposal is probably correct - i.e. the question must be a misprint of \(x^2-5x+1=0\). With this equation you would indeed get a multiple of 5.

  85. TrojanPoem
    • one year ago
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    I like this z+y = 6 from the equation really nice !

  86. asnaseer
    • one year ago
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    thx :)

  87. dan815
    • one year ago
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    oh allsoo dont forgget to not that equation z+y = 6 z^n + y^n is div by 6 only when n is ODD* positive int same thing with the x^2-5x+1

  88. asnaseer
    • one year ago
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    good spot!

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