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Triple primes

  1. Jan 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Here's the problem. We define the triple primes as triples of natural numbers (n,n+2,n+4) for which all three entries are prime. How many triple primes are there? (Hint:mod 3.) (By way of contrast, it is not yet known whether the twin primes-that is, pairs (n,n+2) with both entries prime-form an infinite collection.)


    2. Relevant equations

    see above

    3. The attempt at a solution

    i'm stuck in this problem, don't know where to begin.
     
  2. jcsd
  3. Jan 19, 2009 #2

    Dick

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    You should first ask yourself if there are ANY triple primes. Pay attention to the mod 3 clue.
     
  4. Jan 20, 2009 #3
    i do know that they exist put n=3 and u get (3,5,7). I still dont know how mod 3 would work in this case.
     
  5. Jan 20, 2009 #4

    Dick

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    Oh, yeah. That one. Notice that one of those primes is divisible by three. Write down any 3 consecutive odd numbers and notice that one of them is divisible by 3. Can you give the reason why using a 'mod 3' argument? Then ask yourself how many primes are divisible by 3?
     
  6. Jan 20, 2009 #5
    so if i write (3,5,7) with 3[tex]\equiv0[/tex] mod 3 and 5[tex]\equiv2[/tex] mod 3 and 7[tex]\equiv1[/tex] mod 3 how does that help? and there are no primes which are divisible by 3 unless it would not be a prime number.
     
  7. Jan 20, 2009 #6

    Dick

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    Just do the general case. Take n,n+2,n+4. n mod 3 is either 0 (in which case you are already done),1 or 2. Consider the other two cases and figure out what n+2 and n+4 are.
     
  8. Jan 20, 2009 #7
    n+2 is going to be mod 5 with 0,1,2,3,4. and n+4 is going to be mod 7 with 0,1,2,3,4,5,6. right?
     
  9. Jan 20, 2009 #8
    btw how does n mod 3 = 0 would prove that there is only one triple prime.
     
  10. Jan 20, 2009 #9

    jgens

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    I think what Dick is trying to get you to do is to show that given any natural number n: n, n +2, or n + 4 one of them must be divisible by 3; hence, (3, 5, 7) must be the only set of triple primes.
     
  11. Jan 20, 2009 #10
    i still dont understand the procedure of how i would show that any natural number n,n+2,n+4 must be divisible by 3.
     
  12. Jan 20, 2009 #11
    so for any n, either n, n+2 or n+4 must be divisible by 3 which means either of them should be mod 3?
     
  13. Jan 20, 2009 #12
    Yes, you know that if you have k sequential numbers, one of them will be divisible by k. What's n + 4 mod 3?
     
  14. Jan 20, 2009 #13
    n+4 mod 3 would be 0,1,2.
     
  15. Jan 20, 2009 #14
    I was going to say n + 1 mod 3
     
  16. Jan 20, 2009 #15
    n+1 mod 3 would also give me 0,1,2
     
  17. Jan 20, 2009 #16
    Obviously but the thing I was hoping that you would notice is that you have n, n + 2, n + 4. If we do them mod 3 we get n, n + 1, n + 2 which is THREE sequential numbers so one of them MUST be divisible by 3.
     
  18. Jan 20, 2009 #17
    i still dont understand, what do you mean by a three sequential number?
     
  19. Jan 20, 2009 #18
    By 3 sequential numbers I mean 3 numbers in order, 1, 2 ,3 or 24424, 24425, 24426 or ... n, n + 1, n + 2
     
  20. Jan 20, 2009 #19
    im sorry cause im fairly new to this concept but i have one more thing to ask. i still do not understand how n,n+2,n+4 mod 3 would give you n,n+1,n+2.
     
  21. Jan 20, 2009 #20
    Do you agree that n mod 3 is n mod 3?
    Do you agree that n + 2 mod 3 is n + 2 mod 3?
    Do you agree that n + 4 mod 3 is n + 1 mod 3?
     
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