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estudier

  • 3 years ago

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

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  1. lgbasallote
    • 3 years ago
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    i assume we cannot consider 0

  2. estudier
    • 3 years ago
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    Humph...

  3. estudier
    • 3 years ago
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    Note that any number must be 7k + r with r = 0,1,...6

  4. sauravshakya
    • 3 years ago
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    @estudier i think all numbers can be expressed that way

  5. cwrw238
    • 3 years ago
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    (7k + r)^3 = (7k + r)^2 would that help - i'm clutching at straws to be honest!!

  6. sauravshakya
    • 3 years ago
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    So I think it is 7k+1

  7. estudier
    • 3 years ago
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    ?

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

  9. sauravshakya
    • 3 years ago
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    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
    • 3 years ago
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    So, I think u mean to say 7k+1

  11. estudier
    • 3 years ago
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    I mean every number can be expressed as 7k + r with r = 0,1,...6 Yes, I said that above....

  12. sauravshakya
    • 3 years ago
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    oh.....THAT WAS A HINT

  13. estudier
    • 3 years ago
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    The question does say 7k or 7k +1

  14. NewbieCarrot
    • 3 years ago
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    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
    • 3 years ago
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    @NewbieCarrot Considering certain remainders is a good way to go....

  16. sauravshakya
    • 3 years ago
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    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
    • 3 years ago
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    (7k+r)^2 = 49k^2 + 14kr + r^2 = 7(7k^2+2kr) + r^2

  18. estudier
    • 3 years ago
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    (7k+r)^3 = 343k^3 +147k^2r +21kr^2 +r^3 = 7(49k^3 +21k^2r +3kr^2) +r^3

  19. NewbieCarrot
    • 3 years ago
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    @estudier You got it.

  20. estudier
    • 3 years ago
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    Not done yet....

  21. NewbieCarrot
    • 3 years ago
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    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
    • 3 years ago
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    OK, the reaminders are the same, and.....?

  23. NewbieCarrot
    • 3 years ago
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    The only possible answer is r=0 or 1. And we are done

  24. sara12345
    • 3 years ago
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    n = 7k+r n^2 = 7x + r^2 n^3 = 7y + r^3

  25. sara12345
    • 3 years ago
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    r = [1,6]

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

  27. sara12345
    • 3 years ago
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    0 also

  28. sara12345
    • 3 years ago
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    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
    • 3 years ago
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    Yes, that's right, well done.

  30. estudier
    • 3 years ago
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    Just include r= 0 as well....

  31. sara12345
    • 3 years ago
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    oh yes... .

  32. sauravshakya
    • 3 years ago
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    Is there this rule: If n=x MOD y then, n^a = x^a MOD y

  33. estudier
    • 3 years ago
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    Is there this rule: If n=x MOD y then, n^a = x^a MOD y Yes (r>=1)

  34. estudier
    • 3 years ago
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    (a>=1)

  35. sauravshakya
    • 3 years ago
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    GREAT....... Then I think I proved if n=1,2,3,4,5,6 MOD 7 Then n^6=1 MOD 7

  36. estudier
    • 3 years ago
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    a^(p-1) = 1 mod p (gcd a,p = 1)

  37. estudier
    • 3 years ago
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    p prime

  38. NewbieCarrot
    • 3 years ago
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    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
    • 3 years ago
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    a^(p-1) = 1 mod p (p prime, gcd a,p = 1) (OR a^p = a mod p) Fermat's Little Theorem

  40. estudier
    • 3 years ago
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    @sauravshakya

  41. estudier
    • 3 years ago
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    Indirectly related to the "necklace2 problem....

  42. sauravshakya
    • 3 years ago
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    WHAT!

  43. sauravshakya
    • 3 years ago
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    But how???? That is related to permutation and combination.

  44. estudier
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    (n^p-n)/2p + [n(p+1/2) -n]/2 + n

  47. sauravshakya
    • 3 years ago
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    Did u mean that every colour has only one bead.

  48. estudier
    • 3 years ago
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    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....

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