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More Pythagoras and factoring

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  • David Litchfield
    Just playing with this further there is more note worthy points. Given a number n which is the product of two factors both of which are odd you can derive at
    Message 1 of 1 , Dec 2, 2001
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      Just playing with this further there is more note worthy points.
      Given a number n which is the product of two factors both of which are odd
      you can derive at least 4 pythagoras triangles with each length being whole.
      One of these we've already discussed, one of them is of no major
      important(**) but the other two are:

      21 -> 28 -> 35
      21 -> 72 -> 75

      15 -> 20 -> 25
      15 -> 36 -> 39

      33 -> 44 -> 55
      33 -> 180 -> 183

      35 -> 84 -> 91
      35 -> 120 -> 125

      These dimensions are for such p. triangles but not that when you subtract
      the two dervied lengths from given number they produce the factors
      e.g.
      77 -> 264 -> 275
      77 -> 420 -> 427

      275 - 264 = 11
      427 - 420 = 7

      7 * 11 = 77

      ** the other right angled triangle that can be dervied obtains its side
      lengths from (n^2) /2 + 0.5 and (n^2) /2 - 0.5
      e.g.
      85 -> 3612 -> 3613

      85 ^ 2 = 7225
      7225 /2 = 3612.5

      In terms of "striking it lucky" this helps as we now not just have one
      target to hit but three - all of which can be used to get the factors out.

      Cheers,
      David Litchfield
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