Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

estudier

  • 2 years ago

Show that any number that is both square and cube (eg 64) is of form 7k or 7k +1

  • This Question is Closed
  1. lgbasallote
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i assume we cannot consider 0

  2. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Humph...

  3. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Note that any number must be 7k + r with r = 0,1,...6

  4. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @estudier i think all numbers can be expressed that way

  5. cwrw238
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (7k + r)^3 = (7k + r)^2 would that help - i'm clutching at straws to be honest!!

  6. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So I think it is 7k+1

  7. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ?

  8. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    "(7k + r)^3 = (7k + r)^2" Maybe expand these instead of setting them equal...

  9. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    1^6=7*0+1 2^6=7*9+1 3^6=7*104+1 4^6=7*585+1 I mean every number can be expressed as 7k + r with r = 0,1,...6

  10. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So, I think u mean to say 7k+1

  11. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I mean every number can be expressed as 7k + r with r = 0,1,...6 Yes, I said that above....

  12. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh.....THAT WAS A HINT

  13. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    The question does say 7k or 7k +1

  14. NewbieCarrot
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    For 7k, (7k)^2 (mod 7)= 0^2 (mod 7)=0, (7k)^3 (mod 7)=0^3 (mod 7)=0 For 7k+1, (7k+1)^2 (mod 7)= 1^2 (mod 7)=1, (7k+1)^3 (mod 7)= 1^3 (mod 7)=1 For 7k+2, (7k+2)^2 (mod 7)= 2^2 (mod 7)=4, (7k+2)^3 (mod 7)= 2^3 (mod 7)=1 ......... For 7k+6, (7k+6)^2 (mod 7)= 6^2 (mod 7)=1, (7k+6)^3 (mod 7)= 6^3 (mod 7)=6 for any 7k+r, if the result is r, then it has the same form.

  15. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @NewbieCarrot Considering certain remainders is a good way to go....

  16. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Well x=n^6...... Now, If n=0 MOD 7 then , n^6=0 MOD 7 Thus, x can be expressed as 7k.... Now if n=1,2,3,4,5,6 MOD 7 Then n^6=1 MOD 7 ---> BUT I HAVEN'T PROVED THIS PART.

  17. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    (7k+r)^2 = 49k^2 + 14kr + r^2 = 7(7k^2+2kr) + r^2

  18. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    (7k+r)^3 = 343k^3 +147k^2r +21kr^2 +r^3 = 7(49k^3 +21k^2r +3kr^2) +r^3

  19. NewbieCarrot
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @estudier You got it.

  20. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Not done yet....

  21. NewbieCarrot
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    7(7k^2+2kr) + r^2 7(49k^3 +21k^2r +3kr^2) +r^3 have the same form with 7k+r while r=r^2=r^3 (mod 7)

  22. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    OK, the reaminders are the same, and.....?

  23. NewbieCarrot
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The only possible answer is r=0 or 1. And we are done

  24. sara12345
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    n = 7k+r n^2 = 7x + r^2 n^3 = 7y + r^3

  25. sara12345
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    r = [1,6]

  26. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    "The only possible answer is r=0 or 1. And we are done" True, but a little more explanation would be nice...:-)

  27. sara12345
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    0 also

  28. sara12345
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    let,n = 7k+r (r = 1,2,3,4,5,6) n^2 = 7x + r^2 n^3 = 7y + r^3 plugin r = 1,2,3,4,5,6 and the intersection of remainders gives the form of a number thats both square and cube

  29. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Yes, that's right, well done.

  30. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Just include r= 0 as well....

  31. sara12345
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh yes... .

  32. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Is there this rule: If n=x MOD y then, n^a = x^a MOD y

  33. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Is there this rule: If n=x MOD y then, n^a = x^a MOD y Yes (r>=1)

  34. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    (a>=1)

  35. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    GREAT....... Then I think I proved if n=1,2,3,4,5,6 MOD 7 Then n^6=1 MOD 7

  36. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    a^(p-1) = 1 mod p (gcd a,p = 1)

  37. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    p prime

  38. NewbieCarrot
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Here is my experience, when you see this sentence "is of form 7k or 7k +1" you always should try all 7k+r to get the results. Same as 3k.

  39. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    a^(p-1) = 1 mod p (p prime, gcd a,p = 1) (OR a^p = a mod p) Fermat's Little Theorem

  40. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @sauravshakya

  41. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Indirectly related to the "necklace2 problem....

  42. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    WHAT!

  43. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    But how???? That is related to permutation and combination.

  44. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    n^p-n strings involving 2 or more colours are partitioned into disjoint sets of p strings each set being strings obtained from each other by by a sequence of cycles ie p divides n^p-n which is Fermat's Little Theorem

  45. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Special case of your necklace problem.... Loads of beads of n colours, how many different necklaces of p beads can be made, p is prime?

  46. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    (n^p-n)/2p + [n(p+1/2) -n]/2 + n

  47. sauravshakya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Did u mean that every colour has only one bead.

  48. estudier
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    You make a string of p beads and join the ends, so that's n^p strings. Of those, n strings are beads of 1 colour alone (one for each colour) So n^p-n strings having at least 2 colours etc etc....

  49. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

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.