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Re: Never square

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  • jbrennen
    ... Rewrite this as: N = (4*a+3) * b + (3*a+2) In order for N to be a square, we must have (3*a+2) be a quadratic residue modulo (4*a+3). Note that (3*a+2) is
    Message 1 of 2 , Oct 3, 2002
      --- In primenumbers@y..., "Jon Perry" <perry@g...> wrote:
      > Anyone know why 4ab+3(a+b)+2 is never a square?

      Rewrite this as:

      N = (4*a+3) * b + (3*a+2)

      In order for N to be a square, we must have (3*a+2) be a
      quadratic residue modulo (4*a+3).

      Note that (3*a+2) is equal to -1/4 (modulo 4*a+3).
      [It is the modular solution to 4x+1 == 0]

      -1/4 is a quadratic residue iff -1 is a quadratic residue.

      And finally, -1 is never a quadratic residue modulo 4*a+3;
      this is a basic result of quadratic reciprocity.

      Or for you PARI/GP types who don't care about all of the theory:

      ? for(a=0,99,print(kronecker(3*a+2,4*a+3)))

      ...prints out an endless stream of -1, which gives strong
      evidence that 3*a+2 can never be a quadratic residue
      modulo 4*a+3.
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