Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

estudier Group Title

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

  • one year ago
  • one year ago

  • This Question is Closed
  1. klimenkov Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Does that mean \(2^k-15=a^2, a\in \mathbb N\)?

    • one year ago
  2. estudier Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Looks like a square in integers to me:-)

    • one year ago
  3. estudier Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    OK, naturals if u insist

    • one year ago
  4. estudier Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Number theory type question.....

    • one year ago
  5. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    k=5

    • one year ago
  6. estudier Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    32-15 = 17 is not square

    • one year ago
  7. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    k=4,6 as of now

    • one year ago
  8. estudier Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    That's two....

    • one year ago
  9. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    next one seems a bit far

    • one year ago
  10. estudier Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    "a bit far" ?

    • one year ago
  11. estudier Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    N is a large place:-)

    • one year ago
  12. estudier Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    If u are saying there are no more solutions, saying it is a bit far is insufficient evidence..:-)

    • one year ago
  13. estudier Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Hint Consider odd and even k separately

    • one year ago
  14. shubhamsrg Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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..

    • one year ago
  15. shubhamsrg Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    the k which i used on RHS is different from 2^k

    • one year ago
  16. estudier Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @shubhamsrg That looks nearly right for even k....

    • one year ago
  17. ganeshie8 Group Title
    Best Response
    You've already chosen the best response.
    Medals 4

    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

    • one year ago
  18. shubhamsrg Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    similar to what i was going to write..

    • one year ago
  19. estudier Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    For the odd case, use the fact that any square divided by 3 leaves a remainder of 0 or 1

    • one year ago
  20. ganeshie8 Group Title
    Best Response
    You've already chosen the best response.
    Medals 4

    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

    • one year ago
  21. ganeshie8 Group Title
    Best Response
    You've already chosen the best response.
    Medals 4

    again we get the same solutions for odd case k = {4, 6}

    • one year ago
  22. estudier Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • one year ago
  23. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    • one year ago
    2 Attachments
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.