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Prove that no "prime triplet" exists after 3,5,7

  1. Jun 10, 2014 #1
    So,

    Prime triplet: if p is prime, then p+2 and p+4 are also prime.
    Prove that 3,5,7 is the only prime triplet.

    I have figured out that

    p = 2k+1, k > 1

    then p+2 = 2k + 3 and p + 4 = 2k+5.

    I figure there will be three cases to prove i.e.

    p (prime), p + 2 (prime), p + 4 (not prime)

    p (prime), p + 2 (not prime), p + 4 (prime)

    p (prime), p + 2 (not prime), p +4 (not prime)

    Not sure how to continue. The solution is in the book, but I just want a little hint in the right direction....and I hoped writing it out again would give me some hint. I am just starting with number theory and haven't done many proofs in my life so this feels a little tough.
     
  2. jcsd
  3. Jun 11, 2014 #2

    D H

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    Hint #1: You're thinking too hard. This is an easy problem.
    Hint #2: How many members are in a triplet?
     
  4. Jun 11, 2014 #3

    verty

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    Hint #3: select a few triplets, do you see anything common about them?
     
  5. Jun 11, 2014 #4

    D H

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    And since the OP hasn't replied yet,

    Hint #4: Every such triplet (including 3,5,7) includes one member that is a multiple of what number?
     
  6. Jun 11, 2014 #5
    Your first hint is to look at a group of them and then understand what is special about the anomaly.

    3, 5, 7 works
    5, 7, 9 doesn't
    7, 9, 11 doesn't
    9, 11, 13 doesn't
    11, 13, 15 doesn't

    What do all of these have? How does 3, ,5, 7 get away with it? Can you prove that any triplet will have the same problem?
     
  7. Jun 11, 2014 #6

    Borek

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    Hopefully that's enough hints to get OP started.

    Note how I refrained myself from posting another hint.
     
  8. Jun 11, 2014 #7

    SammyS

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    I took that as a hint to me to also refrain thusly !
     
  9. Jun 11, 2014 #8
    haha thanks so much everyone!
    Every triplet has a multiple of 3. The first triplet survives this because the multiple is 3 itself, and 3 is prime. However, every triplet after the first will contain a multiple of 3 which by the definition of prime numbers means one element is not prime.

    So let's assume p = 3j where j is a positive integer. Then every integer can be formed by 3j + 1 or 3j + 2.

    Then for the first case j = 1 we have p = 3, p + 2 = 5 and p + 4 = 7. No contradiction since we have a prime triplet.

    For the next case 3j + 1 we get p = 3j + 1, p + 2 = 3j + 3 = 3 (j + 1), contradiction since it is a multiple of 3.
    So we can't have a triplets where p is this form.

    Finally, for p = 3j + 2 we get p = 3j + 2, p + 2 = 3j + 4, and p + 4 = 3j + 6 = 3 (j + 2) --> contradiction multiple of 3. Thus, we cannot have prime triplet of this form.

    Therefore no prime triplets exist after 3, 5, 7

    Thanks all!
     
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