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ParthKohli

  • one year ago

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  1. ParthKohli
    • one year ago
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    Alright, so I think freely when I type - so I'll try to solve this myself. I was able to solve the previous one so I'm hoping that I can do this one just as well.

  2. ParthKohli
    • one year ago
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    \[(x+1)^p (x-3)^q = x^n + a_1 x^{n-1} + a_2 x^{n-2}+ \cdots + a_n\]Obviously, \(n = p+q\). We have to find the number of ordered pairs such that \(a_1=a_2\).

  3. ParthKohli
    • one year ago
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    \[1\le p, q \le 1000\]I'll try to think of this in terms of Vieta's Formulas.

  4. ParthKohli
    • one year ago
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    So now, here, we have roots \(-1\) and \(3\) with multiplicities \(p\) and \(q\) respectively. Therefore,\[a_1 = - (-1\cdot p + 3 \cdot q) = p - 3q\]This one was simple. Now let's try to find out \(a_2\), which is the sum taken two at a time.

  5. ParthKohli
    • one year ago
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    Calling @ganeshie8 because I'm feeling quite lonely here.

  6. ganeshie8
    • one year ago
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    \[a_2 = -3\binom{n}{2}\] ?

  7. ParthKohli
    • one year ago
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    Really? Let's see.\[\underbrace{-1, \cdots, -1}_{ p~times}, \underbrace{3, \cdots, 3}_{q ~ times}\]In all, there are \(\binom{n}2\) pairs. Out of those, \(\binom{p}2\) pairs contribute 1 each and \(\binom{q}2\) pairs contribute 9 each. The rest contribute -3 each.

  8. ganeshie8
    • one year ago
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    Ahh right, \(p\ne q\)

  9. ParthKohli
    • one year ago
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    \[a_2 = \binom{p}2\cdot 1 + \binom{q}2\cdot 9 + \left(\binom{p+q}2 - \binom{p}2 - \binom{q}2\right)\cdot (-3) \]

  10. ParthKohli
    • one year ago
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    \[= p - 3q\]

  11. ParthKohli
    • one year ago
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    \[a_2 = \frac{p(p-1) + 9q(q-1)-3\left((p+q)(p+q - 1) - p(p-1) - q(q-1)\right)}{2}\]

  12. ParthKohli
    • one year ago
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    What an ugly expression... let's see if we can size it down a bit.

  13. ganeshie8
    • one year ago
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    \(p,q ~\in~\mathbb{N}\) is it ?

  14. ParthKohli
    • one year ago
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    Yup.

  15. ParthKohli
    • one year ago
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    Expanded and got\[5p^2 - 3q^2 - 3 p - 3q + 6pq = 0\]This is some kind of a conic section, ain't it?

  16. ganeshie8
    • one year ago
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    \[8p+24q+1 = (2p-6q+1)^2\]

  17. ParthKohli
    • one year ago
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    Wow, that looks better. How do we find integral solutions for this?

  18. ParthKohli
    • one year ago
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    Like it means that \(8p + 24q + 1\) is a perfect square and there aren't many perfect squares in the range 1-1000, but still there are a lot of them and we can't keep track. You're the number-theory guy here.

  19. ganeshie8
    • one year ago
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    wait, do we need to find the ordered pairs (p, q) in the range 1-1000 ?

  20. ParthKohli
    • one year ago
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    Yeah.

  21. ParthKohli
    • one year ago
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    Oh, so we need to look at perfect squares in the form \(8k + 1\) right?

  22. ParthKohli
    • one year ago
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    Well, unfortunately, that means all squares of odd numbers. Haha.

  23. ganeshie8
    • one year ago
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    I think so because the left hand side is 8k+1

  24. ganeshie8
    • one year ago
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    oh yeah that doesn't help much @mukushla

  25. ParthKohli
    • one year ago
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    How did you come up with that form, ganeshie?

  26. anonymous
    • one year ago
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    Nice question, It's late here guys, I'll come back to this later :) good night!

  27. ParthKohli
    • one year ago
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    BTW, I found this one on Brilliant too...

  28. ParthKohli
    • one year ago
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    And good night!

  29. anonymous
    • one year ago
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    It must be\[p^2 -6pq +9q^2 - 3 p - 3q = 0\]

  30. anonymous
    • one year ago
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    solve for \(q\)\[q=\frac{2p+1 \pm \sqrt{16p+1}}{6}\]

  31. ganeshie8
    • one year ago
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    does that mean 16p+1 must be a perfect square

  32. anonymous
    • one year ago
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    right

  33. ganeshie8
    • one year ago
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    Nice! there won't be too many perfect squares of this form is there a nice way to account for divisibility by 6

  34. ganeshie8
    • one year ago
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    \[j^2 \equiv 1\pmod{16} \implies j\equiv \pm 1, ~\pm 7 \pmod{16}\]

  35. anonymous
    • one year ago
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    let \(16p+1=n^2\) which gives\[16p=(m-1)(m+1)\]now \(m\) must be an odd number, say \(2k+1\) \[4p=k(k+1)\]how about \(k\)? \(k\) has the forms of \(4l\) or \(4l+3\) only.

  36. anonymous
    • one year ago
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    * \(16p+1=m^2\)

  37. ganeshie8
    • one year ago
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    does that mean \(p\) has to be an even triangular number ?

  38. anonymous
    • one year ago
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    I think yes, let's finish this first

  39. ganeshie8
    • one year ago
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    \[4p=k(k+1) \implies p =\dfrac{k(k+1)}{2*2}=\dfrac{T_k}{2}\]

  40. anonymous
    • one year ago
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    aha, counting triangular numbers is easy ha?

  41. anonymous
    • one year ago
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    we'll work on that too, from mine you get \(p=l(4l+1)\) or \(p=(l+1)(4l+3)\)

  42. ganeshie8
    • one year ago
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    that looks a lot better! i see where it is going, we work q also similarly right

  43. anonymous
    • one year ago
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    we have the quadratic formula, we just need to plug p in terms of l to get q

  44. anonymous
    • one year ago
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    guys dont want to disteurb this meeting after u are done can u explain this to me from point blank point thanks

  45. ganeshie8
    • one year ago
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    \[q=\frac{2p+1 \pm \sqrt{16p+1}}{6} = \dfrac{2p+1\pm (8l+7)}{6}\] ?

  46. anonymous
    • one year ago
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    right

  47. ganeshie8
    • one year ago
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    subing p also \[q=\frac{2p+1 \pm \sqrt{16p+1}}{6} = \dfrac{2(l+1)(4l+3)+1\pm (8l+7)}{6}\]

  48. anonymous
    • one year ago
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    ok we have some pairs of \(p, q\) in terms of \(l\)\[l(4l+1), \frac{l(4l-3)}{3} \\ l(4l+1), \frac{(l+1)(4l+1)}{3}\]

  49. dan815
    • one year ago
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    |dw:1435058536781:dw|

  50. dan815
    • one year ago
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    im prolly really behind was this done already

  51. ganeshie8
    • one year ago
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    right, 4 cases we can start counting i guess... shouldn't be a problem xD

  52. anonymous
    • one year ago
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    \[(l+1)(4l+3), \frac{(l+1)(4l+7)}{3} \\ (l+1)(4l+3), \frac{4l^2}{3}+l\]

  53. anonymous
    • one year ago
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    this is not good xD

  54. ganeshie8
    • one year ago
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    ok we have some pairs of \(p, q\) in terms of \(l\) case1,2 \[l(4l+1), \frac{l(4l-3)}{3} \\ l(4l+1), \frac{(l+1)(4l+1)}{3}\] case3,4 \[(l+1)(4l+3), \frac{(l+1)(4l+7)}{3} \\ (l+1)(4l+3), \frac{4l^2}{3}+l\]

  55. dan815
    • one year ago
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    why did u guys change to L =[

  56. dan815
    • one year ago
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    |dw:1435059008705:dw|

  57. ikram002p
    • one year ago
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    i dont know even what u guys trying to solve i'd like to start over

  58. dan815
    • one year ago
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    6 nice equations i think 3 of 4 are pretty trivial

  59. ganeshie8
    • one year ago
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    @dan815 equating your expressions for \(a_1\) and \(a_2\), we end up with below quadratic : \[p^2 -6pq +9q^2 - 3 p - 3q = 0\]

  60. ganeshie8
    • one year ago
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    then use quadratic formula to isolate \(q\)

  61. anonymous
    • one year ago
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    @ganeshie8 wait, I found a better method, this is very short

  62. ganeshie8
    • one year ago
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    ok waiting xD

  63. anonymous
    • one year ago
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    \[(p-3q)^2=3(p+q)\]right?

  64. ikram002p
    • one year ago
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    eh i feel like pluto being isolated from this mathematicians council -.-

  65. ganeshie8
    • one year ago
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    wow ok so \(3\mid (p-3q)^2 \implies 3\mid(p-3q)\)

  66. anonymous
    • one year ago
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    \(p+q=3m^2\) for \(m \in \mathbb{N}\) and \(p-3q=\pm3 m\) solve for \(p\) and \(q\) in terms of \(m\) ;-)

  67. dan815
    • one year ago
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    there are just 2?

  68. dan815
    • one year ago
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    oh theres 1 more qeuation possible

  69. ganeshie8
    • one year ago
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    \[a_1 = p - 3q\tag{1}\] \[a_2 = \binom{p}2\cdot 1 + \binom{q}2\cdot 9 + \left(\binom{p+q}2 - \binom{p}2 - \binom{q}2\right)\cdot (-3)\tag{2} \] @dan815 how did u get ur cases from above two equations

  70. dan815
    • one year ago
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    i kept the cases separate

  71. ganeshie8
    • one year ago
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    |dw:1435060009607:dw|

  72. anonymous
    • one year ago
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    guys I have found the final answer as 19, see if this is right or not

  73. dan815
    • one year ago
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    |dw:1435060528572:dw|

  74. ganeshie8
    • one year ago
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    19 pairs by both methods!

  75. ganeshie8
    • one year ago
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    okie

  76. ikram002p
    • one year ago
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    well that was very exhausted question for me to follow, but indeed unforgettable for a last question >.<

  77. dan815
    • one year ago
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    I dont see how else u can get it down to only 19

  78. dan815
    • one year ago
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    lemme think a little more

  79. dan815
    • one year ago
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    ah omg i found my mistake xD

  80. ganeshie8
    • one year ago
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    nice :) how many ordered pairs are u getting

  81. dan815
    • one year ago
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    hey I was afk, i get the same equations i left it there

  82. dan815
    • one year ago
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    |dw:1435075471169:dw|

  83. dan815
    • one year ago
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    hmm integer solutions to this eh

  84. dan815
    • one year ago
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    |dw:1435075773197:dw|

  85. dan815
    • one year ago
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    |dw:1435075838340:dw|

  86. dan815
    • one year ago
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    can set a max bound to this now

  87. ganeshie8
    • one year ago
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    yes : \(p+q=3m^2\) for \(m \in \mathbb{N}\) and \(p-3q=\pm3 m\) solve for \(p\) and \(q\) in terms of \(m\) ;-)

  88. dan815
    • one year ago
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    how come u got p+q = 3m^2?

  89. dan815
    • one year ago
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    is it because u know that p+q is an int and we get (p-3q)^2/3 = p+q

  90. ganeshie8
    • one year ago
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    |dw:1435076243816:dw|

  91. dan815
    • one year ago
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    right okay so lets say (p-3q)^2 = 3*m^2 p-3q=+/- 3m

  92. ganeshie8
    • one year ago
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    \(3\mid (p-3q)^2 \implies 3\mid(p-3q) \implies p-3q = 3m\)

  93. ganeshie8
    • one year ago
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    |dw:1435076343662:dw|