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Primes of the form a^2-b^2

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  • sopadeajo2001
    Primes can be represented as a^2-b^2=(a+b)(a-b) if we choose a-b=1 and a+b=p so a=(p+1)/2 and b=(p-1)/2 so a and b are consecutive integers one odd and another
    Message 1 of 1 , Jun 17, 2009
      Primes can be represented as a^2-b^2=(a+b)(a-b) if we choose a-b=1 and a+b=p so a=(p+1)/2 and b=(p-1)/2 so a and b are consecutive integers one odd and another even and so all odd numbers (and thus primes too) can be represented as difference of consecutive squares:

      1^2-0^2=1
      2^2-1^2=3
      3^2-2^2=5
      4^2-3^2=7

      If we sum them we get:
      1+3=2^2
      1+3+5=3^2
      1+3+5+7=4^2
      The sum of the first n odd integers (beginning with 1) is n^2 (1)

      If 1+....+2n-1=n^2 then 1+....+2n-1+2n+1=n^2+2n+1=(n+1)^2
      (1) is proved by induction

      And any prime can be represented as a difference of squares ((p+1)/2)^2-((p-1)/2)^2 while only 1 mod 4 primes can be represented as a sum of squares (all in N)
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