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Prime Numbers

  • Thread starter Pandaren
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  • #1
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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.
 
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Answers and Replies

  • #2
Hurkyl
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Try factoring.
 
  • #3
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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:
 
  • #4
Hurkyl
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You've found 2 factors, in terms of v.
Can you find all factors, in terms of x?
 
  • #5
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I dont understand the "all factors of x" part
you mean [tex](x - 1)(x + 1) = v^3[/tex] ?
 
  • #6
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The only x that seem to work is x = 2
 
  • #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...
 
  • #8
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Hurkyl said:
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...
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]?
 
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  • #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.
 
  • #10
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Ohh I understand now thanks for your help
 

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