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Prime Numbers

  1. Oct 16, 2004 #1
    Hey everyone, I need help on this problem:

    Find all prime numbers [tex]x[/tex] such that [tex]x^2 = v^3 + 1[/tex] for some integer [tex]v[/tex].

    Thanks a lot for your help, i appreciated.
     
    Last edited: Oct 16, 2004
  2. jcsd
  3. Oct 16, 2004 #2

    Hurkyl

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    Try factoring.
     
  4. Oct 16, 2004 #3
    After I factored the right side, i got [tex]x^2 = (v + 1)(v^2 - v + 1)[/tex] I dont see what to do next :frown:

    Thank you for your help :smile:
     
  5. Oct 16, 2004 #4

    Hurkyl

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    You've found 2 factors, in terms of v.
    Can you find all factors, in terms of x?
     
  6. Oct 16, 2004 #5
    I dont understand the "all factors of x" part
    you mean [tex](x - 1)(x + 1) = v^3[/tex] ?
     
  7. Oct 16, 2004 #6
    The only x that seem to work is x = 2
     
  8. Oct 16, 2004 #7

    Hurkyl

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    Well, x^2 is the square of the prime, x, right? You know the prime factorization of x^2, so you can write down all of its factors.


    Since v+1 is a factor, v+1 must be equal to one of the numbers on that list...
     
  9. Oct 16, 2004 #8
    you mean if 2 is a prime, then [tex]2 = x[/tex], [tex]2^2 = x^2[/tex]
    Then [tex]2^2 = (v + 1)(v^2 - v + 1)[/tex] Or [tex]2 * 2 = (v + 1)(v^2 - v + 1)[/tex]
    Therefore [tex]2 = v + 1[/tex] and [tex]2 = v^2 - v + 1[/tex]?
     
    Last edited: Oct 16, 2004
  10. Oct 17, 2004 #9

    Hurkyl

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    You know that (v+1)(v^2-v+1) is a factorization of x^2 into two factors, and you know that x * x is a factorization of x^2 into two factors, but you don't know that they're the same factorization.

    However, you can write down all of the ways to factor x^2 into two factors, and you know that (v+1)(v^2-v+1) is going to be one of them.
     
  11. Oct 17, 2004 #10
    Ohh I understand now thanks for your help
     
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