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Hypotenuse as required

  1. Feb 21, 2016 #1
    I wonder if I have found the most efficient way of finding any pythagoras triple with a given hypotenuse that has not been published?
    My method uses a trial and error search, but with the limits set to minimize the number of trials as follows:
    Try odd values X within the limits of int(C^0.5) +1 to int(2.C)^0.5 inclusive such that
    2.C - X^2 = Y^2. C being the required hypotenuse. {This is always possible and in a number of ways depending on the number (at least one) of prime factors of form 4n+1 in C.}
    Then the required value of one side of the triangle is given by X.Y
    Example C= 65
    lower limit = 9 upper limit = 11
    X = 11, Y = 3 or X = 9 Y = 7

    33^2 + 56^2 = 65^2 or 63^2 + 16^2 = 65^2

    These are primitive solutions, but non primitive ones can be found by applying the above method to C = 5 or 17 and then multiplying the three sides by 17 or 5.
     
  2. jcsd
  3. Feb 21, 2016 #2

    Mark44

    Staff: Mentor

    I moved this thread from the textbooks section, as it seemed more related to the mathematics and very little to textbooks.
     
  4. Feb 21, 2016 #3

    fresh_42

    Staff: Mentor

    Your examples can all be found by the equation ##(u^2-v^2)^2 + (2uv)^2 = (u^2+v^2)^2## which is well known.
    I can't remember and haven't looked up whether all triples can be found this way. Nevertheless, your's can be found, e.g. ##65^2=8^2+1^2=7^2+4^2##.
     
  5. Feb 21, 2016 #4

    mathman

    User Avatar
    Science Advisor
    Gold Member

    This gives all triples.
     
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