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estudier Group Title

2^k -15 is square. Find all k (integers)

  • 2 years ago
  • 2 years ago

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  1. klimenkov Group Title
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    Does that mean \(2^k-15=a^2, a\in \mathbb N\)?

    • 2 years ago
  2. estudier Group Title
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    Looks like a square in integers to me:-)

    • 2 years ago
  3. estudier Group Title
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    OK, naturals if u insist

    • 2 years ago
  4. estudier Group Title
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    Number theory type question.....

    • 2 years ago
  5. sauravshakya Group Title
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    k=5

    • 2 years ago
  6. estudier Group Title
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    32-15 = 17 is not square

    • 2 years ago
  7. hartnn Group Title
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    k=4,6 as of now

    • 2 years ago
  8. estudier Group Title
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    That's two....

    • 2 years ago
  9. hartnn Group Title
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    next one seems a bit far

    • 2 years ago
  10. estudier Group Title
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    "a bit far" ?

    • 2 years ago
  11. estudier Group Title
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    N is a large place:-)

    • 2 years ago
  12. estudier Group Title
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    If u are saying there are no more solutions, saying it is a bit far is insufficient evidence..:-)

    • 2 years ago
  13. estudier Group Title
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    Hint Consider odd and even k separately

    • 2 years ago
  14. shubhamsrg Group Title
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    maybe i got somewhere 2^k = m^2 or 2m^2 for 2^k = m^2 m^2 - 15 = k^2 (m+k)(m-k) = 1*3*5 on comparison, m+k = 5 and m-k =3 =>m=4 -->1st solution again on comparison, m+k =15 m-k =1 m=8 -->2nd solution just have to see through 2m^2 now..

    • 2 years ago
  15. shubhamsrg Group Title
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    the k which i used on RHS is different from 2^k

    • 2 years ago
  16. estudier Group Title
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    @shubhamsrg That looks nearly right for even k....

    • 2 years ago
  17. ganeshie8 Group Title
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    2^k-15 = p^2 2^k-p^2 = 15 lets say k is even, k = 2t 2^2t - p^2 = 15 (2^t + p)(2^t-p) = 15 solving, we get few solutions lets say k is off, k = 2t+1 2^2t+1 - p^2 = 15 (2^2t+1/2 + p)(2^2t+1/2 - p) = 15 solving we may not get any solutions on cursory check.. only k = 4,6 are the solutions

    • 2 years ago
  18. shubhamsrg Group Title
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    similar to what i was going to write..

    • 2 years ago
  19. estudier Group Title
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    For the odd case, use the fact that any square divided by 3 leaves a remainder of 0 or 1

    • 2 years ago
  20. ganeshie8 Group Title
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    for the even case, k = {4,6} for the odd case : 2.2^2t+1/2 = 16 2.2^2t+1/2 = 8 we need to solve above both

    • 2 years ago
  21. ganeshie8 Group Title
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    again we get the same solutions for odd case k = {4, 6}

    • 2 years ago
  22. estudier Group Title
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    K, for the odd you can also say that 2^k has remainder 2 on division by 3 and then so does 2^k -15 which is not 0 or 1 so there are no solutions when k is odd

    • 2 years ago
  23. sauravshakya Group Title
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    • 2 years ago
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