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estudier

  • 3 years ago

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

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  1. klimenkov
    • 3 years ago
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    Does that mean \(2^k-15=a^2, a\in \mathbb N\)?

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

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

  4. estudier
    • 3 years ago
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    Number theory type question.....

  5. sauravshakya
    • 3 years ago
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    k=5

  6. estudier
    • 3 years ago
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    32-15 = 17 is not square

  7. hartnn
    • 3 years ago
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    k=4,6 as of now

  8. estudier
    • 3 years ago
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    That's two....

  9. hartnn
    • 3 years ago
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    next one seems a bit far

  10. estudier
    • 3 years ago
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    "a bit far" ?

  11. estudier
    • 3 years ago
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    N is a large place:-)

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

  13. estudier
    • 3 years ago
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    Hint Consider odd and even k separately

  14. shubhamsrg
    • 3 years ago
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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..

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

  16. estudier
    • 3 years ago
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    @shubhamsrg That looks nearly right for even k....

  17. ganeshie8
    • 3 years ago
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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

  18. shubhamsrg
    • 3 years ago
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    similar to what i was going to write..

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

  20. ganeshie8
    • 3 years ago
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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

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

  22. estudier
    • 3 years ago
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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

  23. sauravshakya
    • 3 years ago
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