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RE A question about products of primes

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  • Jose Ramón Brox
    ... From: bill2math A. I am looking for a solution that also includes n as a factor if n is prime, not just the primes below n.
    Message 1 of 3 , Oct 29, 2005
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      ----- Original Message -----
      From: "bill2math" <bill2math@...>

      A. I am looking for a solution that also includes n as a factor if
      n is prime, not just the primes below n.


      Hi again, Bill:

      If n is a prime p, then by your hipothesis, n+1 must be (n-2)#. But by Bertrand's
      Postulate, if p>8, then (n-2)# >= (n-2)/2*(n-2)/4 = (n-2)^2 / 4
      Therefore, it must hold that n+1 >= (n-2)^2 / 4 --> 4(n+1) > = n^2 -4n + 4 --> n^2 -8n <=
      0 --> (n-8) <= 0 --> n <= 8. If n>8, then it's not possible.

      If n<=8, then we can do a list and find that the only solutions are n=2,3,5.


      B. In statement 2.a) you say "If p>(n+1)/2 then p can't divide
      neither n nor n+1." Not true given my comments in "A" above. If
      p=n, then of course p divides n.


      Yeah, you are right. I overlooked the case where n itself is prime. Never mind, it is
      resolved like in A above ;-)


      C. According to Mathworld, Bertrand's postulate states that there
      is always a prime p such that k<p<2k. Therefore, the equalities
      (p=n/2 in statements 1.b) and (p=(n+1)/2 in statement 2.b) are

      Well, if a theorem says that A < B < C then it's nothing wrong in saying that A <= B <= C
      holds. But you are right, it was pointless to check the cases 1.b) and 2.b), I used
      Bertrand's Postulate by memory and I didn't remember if it was strict or not in his
      restrictions, so I used the less restricted form.


      D. It is not clear to me how you arrive at your final statement
      "and this is only affordable if p=3 and n=5, otherwise n is bigger
      than n+1." Why is it only affordable if p=3 and n=5?


      Because of your appreciation, we don't need more the point 2.b), and that statement was
      within it, so it was a waste. Nevertheless, I said it wrong, because n=2 and n=3 were also
      possible solutions. But they were all lower than 6, and your problem was about n>=6.

      So your claim is solved in the negative, if n>=6 we can always find a prime lower than n+1
      that is not a divisor of n(n+1). Said otherwise, if n>=6 then n(n+1) is never a multiple
      of n#.

      I hope it is all right now.

      Regards. Jose Brox
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