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RE Perfect square problem

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  • Jose Ramón Brox
    A + x^2 = y^2 A = y^2 - x^2 A = (y+x)(y-x) It depends on the number of divisions of A in two divisors. There is a finite number of solutions to the equation.
    Message 1 of 1 , Oct 21, 2004
      A + x^2 = y^2

      A = y^2 - x^2

      A = (y+x)(y-x)

      It depends on the number of divisions of A in two divisors. There is a
      finite number of solutions to the equation. You can't find an y for every x.
      You must equal y+x to the biggest divisor F and y-x to the smallest f, where
      F*f = A. Then y = (F+f) / 2 and x = (F-f) / 2. You will get a solution only
      for those divisors whose sum is even, so the numbers n = 2*k where MCD(k,2)
      = 1 don't have a solution.

      For example:

      A = 12 --> A = 12 * 1 = 6 * 2 = 4 * 3, only 6 + 2 is even.

      x+y = 6
      y-x = 2

      x = 2, y = 4

      12 + 2^2 = 4^2 --> 12 + 4 = 16 Ok.

      I hope it's useful.

      Jose Brox



      ----- Original Message -----
      From: "Guna" <gunaseelapandian@...>
      To: <primenumbers@yahoogroups.com>
      Sent: Thursday, October 21, 2004 1:42 PM
      Subject: [PrimeNumbers] perfect square problem


      >
      > Is there a way mathematically to find out two numbers such that one is a
      square (I mean whole) when a known number is added with the second's square
      ?
      >
      > for example let A be a number that is known
      >
      > now we add (x)2 to A and the result is a perfect square y such that the
      square root of y is lets say r that does not have a decimal part.
      >
      > I am not a mathematician but am working on primes for almost 3 months now.
      This is really fascinating.
      >
      > Guna
      >
      >
      >
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