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Marc's conjecture: Primes in consecutive arithmetic series.

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  • Kermit Rose
    Marc Persuy, conjectured that For each prime number p, it exists x pairs (x 0) of smaller prime numbers q r such that p-q=q-r , which can be expressed by: p is
    Message 1 of 1 , Dec 21, 2005
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      Marc Persuy,
      conjectured
      that
      For each prime number p, it exists x pairs (x>0) of smaller prime numbers q
      r such that
      p-q=q-r , which can be expressed by: p is symmetric from r through q.


      At first I thought this not significant because this is a property of odd
      integers

      For any two odd integers p, q, where q is less than p,

      r = 2 q - p is odd,

      and p - q = q - p.



      Marc Persuy conjectures that his property of odd integers partially extends
      to
      the prime number subset of the odd integers.


      He conjectures that if p is prime then

      there exist primes r and q with r < q such that p - q = q - r.


      It's easily proven that if p is an odd prime, then

      there exist odd numbers r and q with r < q such that p - q = q - r.

      For take q any odd number between (p+1)/2 and p; define r = 2 q - p.

      So now we can almost prove Marc's conjecture.

      We can prove that if p is an odd prime then there exist a prime q and and
      odd positive integer r < q

      such that q - r = p - q.

      It has been proven that between every integer and its double, there is at
      least one prime.

      Thus if p is an odd prime > 5
      Thus there is at least one odd prime between (p+1)/2 and p.

      Let q be such a prime.

      Then define r = 2 q - p.

      Then q - r = p - q.


      Now can we go one more step and prove Marc's conjecture?


      We've selected the prime p, and a prime q between (p+1)/2 and p.

      How can we find which odd numbers r = 2 q - p are prime?


      We proceed as follows:

      We seek primes r, q, p such that

      q - r = p - q.

      Define k = q - r.

      Then this relationship becomes

      q = r + k or r = q - k.


      p = q + k = r + 2 k



      or, in terms of p,

      q = p - k,

      r = p - 2 k = q - k


      Thus marc's conjecture is equivalent to the conjecture that

      for every prime p, it's possible to find primes q and r such that

      for some positive integer k,

      q = p - k and r = q - k.


      The following has a trivial proof.

      If q is any odd positive integer, and k is any positive integer,

      and

      if the equation

      m n + k m + n = q has only the solutions

      {m = 0, n = q } and { m = q+k-1, n = 1 - k } in integral m and n,

      and

      m n + k m - n = q

      also has only the solutions

      { m = 2, n = q - 2 k} and { m = q - k + 1, n = 1 - k}



      then both

      p = q + k

      and

      r = q - k

      are prime.


      So to complete the proof of Marc's conjecture, we need only prove that

      given a prime q between (p+1)/2 and p,

      there exist an integer k such that

      that each of the equations

      m n + k m + n = q

      and

      m n + k m - n = q

      have exactly two integral solutions.





      In fact, since we have already specified p as prime, we need only look at
      the

      second equation.

      If p and q are prime, with p > q,

      and

      if the positive integer k = p - q has the property that

      the equation

      m n + k m - n has exactly two integral solutions,

      then r = p - q is prime.
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