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RE: [PrimeNumbers] short proofs

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  • Jon Perry
    Is n!-1 ever a square? There must exist some k such that n!+/-k is never square, other than k=0. Jon Perry perry@globalnet.co.uk
    Message 1 of 7 , Apr 20, 2003
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      Is n!-1 ever a square?

      There must exist some k such that n!+/-k is never square, other than k=0.

      Jon Perry
      perry@...
      http://www.users.globalnet.co.uk/~perry/maths/
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    • Jon Perry
      If the Goldbach Conjecture holds, then for every n there exist p,q primes that verify p+q = 2·n -- ... 2) p q =n (could be p=q=n) Conclusion: n
      Message 2 of 7 , Apr 20, 2003
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        'If the Goldbach Conjecture holds, then for every n there exist p,q primes
        that verify p+q = 2·n -->

        ---> 1) p, q < 2·n
        2) p<=n --> q>=n (could be p=q=n)

        Conclusion: n <= q < 2·n --> Proof for Bertrand's postulate: between n and
        2n there is always a prime.'

        I think everyone who has ever attempted to solve GC has spotted this.

        Jon Perry
        perry@...
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      • xyzenogenicyx
        everyone please send in a cool short proof based on primes. one of my favorites is , the factorial of n with n 1 can never be a square... using bertrands
        Message 3 of 7 , Apr 21, 2003
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          everyone please send in a cool short proof based on primes.


          one of my favorites is , the factorial of n with n > 1 can never be a
          square... using bertrands postulate there is always a prime between
          k and 2k. From this it follows that there is a prime that divides n!
          but whose square does not. because in the prime
          factorization of a square all exponents are even, thus n! cannot be a
          perfect square.
        • Paul Leyland
          ... Here is an old favourite: there is an infinite number of primes. Proof: Define F_n = 2^(2^n) + 1 where n is an integer = 1. Now F_n - 2 = 2^(2^n) - 1 =
          Message 4 of 7 , Apr 21, 2003
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            > everyone please send in a cool short proof based on primes.

            Here is an old favourite: there is an infinite number of primes.

            Proof:

            Define F_n = 2^(2^n) + 1 where n is an integer >= 1.

            Now F_n - 2 = 2^(2^n) - 1 = (2^(2^n-1) + 1) * (2^(2^n-1) - 1) by the difference of squares rule.

            Hence F_n - 2 = F_{2n-1} * (F_{2n-1} - 2))

            So F_n leaves a remainder 2 when divided by F_{n-1} and so F_n is not divisible by any F_m when m<n and so the prime factorization of F_n does not include any F_m with m<n

            As n may take on an infinity of values, there are an infinite number of primes.


            Paul
          • Jose Ramón Brox
            Here comes a little proof of mine (oh, my big ego :P): If the Goldbach Conjecture holds, then for every n there exist p,q primes that verify p+q = 2·n -- ...
            Message 5 of 7 , Apr 21, 2003
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              Here comes a little proof of mine (oh, my big ego :P):

              If the Goldbach Conjecture holds, then for every n there exist p,q primes that verify p+q = 2·n -->

              ---> 1) p, q < 2·n
              2) p<=n --> q>=n (could be p=q=n)

              Conclusion: n <= q < 2·n --> Proof for Bertrand's postulate: between n and 2n there is always a prime.

              Can anyone say if this was observed yet for any author? (I don't like to say that is my proof if it isn't).

              Jose Brox
              http://espanol.groups.yahoo.com/group/Telecomunicacion/

              ----- Original Message -----
              From: xyzenogenicyx
              To: primenumbers@yahoogroups.com
              Sent: Monday, April 21, 2003 8:27 PM
              Subject: [PrimeNumbers] short proofs



              everyone please send in a cool short proof based on primes.


              one of my favorites is , the factorial of n with n > 1 can never be a
              square... using bertrands postulate there is always a prime between
              k and 2k. From this it follows that there is a prime that divides n!
              but whose square does not. because in the prime
              factorization of a square all exponents are even, thus n! cannot be a
              perfect square.



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            • Jose Ramón Brox
              ... I also think so, this is the reason I ask for the references
              Message 6 of 7 , Apr 21, 2003
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                Jon says:
                > I think everyone who has ever attempted to solve GC has spotted this.

                I also think so, this is the reason I ask for the references
              • Jose Ramón Brox
                What happened to this thread? I enjoyed it. Would be any web publications of the proof recolected? Are any more proofs wanted? Jose Brox ... From:
                Message 7 of 7 , May 4 2:19 PM
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                  What happened to this thread? I enjoyed it.

                  Would be any web "publications" of the proof recolected?

                  Are any more proofs wanted?

                  Jose Brox


                  ----- Original Message -----
                  From: xyzenogenicyx
                  To: primenumbers@yahoogroups.com
                  Sent: Monday, April 21, 2003 8:27 PM
                  Subject: [PrimeNumbers] short proofs



                  everyone please send in a cool short proof based on primes.


                  one of my favorites is , the factorial of n with n > 1 can never be a
                  square... using bertrands postulate there is always a prime between
                  k and 2k. From this it follows that there is a prime that divides n!
                  but whose square does not. because in the prime
                  factorization of a square all exponents are even, thus n! cannot be a
                  perfect square.



                  Yahoo! Groups Sponsor



                  Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
                  The Prime Pages : http://www.primepages.org/



                  Your use of Yahoo! Groups is subject to the Yahoo! Terms of Service.



                  [Non-text portions of this message have been removed]
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