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Number theory problem

  1. Jul 8, 2010 #1
    1. The problem statement, all variables and given/known data

    If k is a prime number
    find all k that satisfy k²=n³+1
    n is not a prime number

    2. Relevant equations

    I really have no idea, use any suitable one

    3. The attempt at a solution

    all prime numbers are odd except 2.
    n must be positive natural number
    n³ = k² -1 = (k-1)(k+1)

    any help is appreciated
     
  2. jcsd
  3. Jul 8, 2010 #2
    Two things you need to know to solve this problem:
    1) If k is prime, then k^2 can only be divided by 4 numbers, 2 of which are k^2 and 1.
    2) a^3 + b^3 = (a^2-ab+b^2)(a+b)

    If you need more clues, ask, but post anything you did with these clues:)

    -Tusike
     
  4. Jul 8, 2010 #3
    thanks I'll try these hints
     
  5. Jul 8, 2010 #4
    I tried with these two clues,
    k^2 = (n+1)(n^2 -n+1)
    k = sqrt [(n+1)(n^2 +n+1)]
    and from clue (1) k^2 can be divided by k^2 , 1, k and -k
    but I don't know how to connect the information together to get the answer :(
     
  6. Jul 8, 2010 #5
    Ugh sorry, I said something wrong
    k^2 can only be divided by 3 numbers, not 4 (I was thinking how it can be written down two ways as a multiple of two numbers...) . So what does that say about (n+1) and (n^2 -n + 1)?
     
  7. Jul 8, 2010 #6
    k^2= (n+1) [ n(n-1) +1]
    that is
    k^2 = n (n+1)(n-1) + (n+1)
    right?
     
  8. Jul 8, 2010 #7
    You don't need to solve that equation just yet, only see how it can happen.
    On the left side, you have k^2, and on the right side you have a multiple of two whole numbers, (n+1) and (n^2 - n + 1).
    You know that k^2 can either be written as 1*k^2, k^2*1, or sqr(k^2)*sqr(k^2). These are also multiples of two whole numbers! So just pair them up:) You should get n=2, and from then k=3. 1*k^2 and k^2 * 1 won't lead to anything worth noticing.

    EDIT: sqr(k^2), not sqr(k), sorry...
     
  9. Jul 8, 2010 #8
    thaaaaaaaaaaaaaank you!
    please check the steps
    (n+1) = k
    (n^2 -n +1) = k
    n^2 -n +1 = n+1
    n(n-2)=0 n=0 , k=1 (not prime)
    n=2 , k=3
     
    Last edited: Jul 8, 2010
  10. Jul 8, 2010 #9
    Yes, that is correct. Now I don't know about how "precise" this proof has to be, but if has to be really precise, you might want to examine the following two possibilities as well:
    (1)
    (n+1) = 1 and (n^2 - n + 1) = k^2

    (2)
    (n+1) = k^2 and (n^2 - n + 1) = 1

    From the first you'll get that n = 0 and k = 1, which of course isn't prime, so that's not a good solution.
    From the second, you'll get that n = 1, and k=sqr(2), which again isn't prime. However, you couldn't of known that you won't get valid solutions from these two, so I think they need to be added to the proof as well to make it complete.
     
  11. Jul 8, 2010 #10
    Thank you very much,Tusike
    ^_^
     
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