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RE: [PrimeNumbers] Pythagoras theorem

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  • cino hilliard
    Hi Peter, Not 100% on topic but close enough for a response from me. Chances are number fiddlers, hypothesizers and number enthusiasts make this discovery
    Message 1 of 3 , Jul 14, 2009
      Hi Peter,



      Not 100% on topic but close enough for a response from me.



      Chances are number fiddlers, hypothesizers and number enthusiasts make this

      discovery every day. I pounced upon it messing around with x^n+y^n = z^n

      in my yout back in the late 50's. At that time I assumed your result were the
      only solutions.



      What I should have done was look more closely at the numbers at that time.



      n x y z

      1 3 4 5

      2 5 12 13

      3 7 24 25

      4 9 40 41

      5 11 60 61

      ....



      Indeed, 3^2+4^2=5^2, 5^2 + 12^2 = 13^2, etc

      Notice that x is the set of odd numbers. Then notice that

      n x

      1 3 = 4-1

      2 5 = 9-4

      3 7 = 16-9

      4 9 = 25-16

      ...

      Yep looks like x is the difference between 2 squares. Now

      n z

      1 5 = 4+1

      2 13 = 9+4

      3 25 = 16+9

      4 41 = 25+16

      ...

      Yep looks like z is the sum of 2 squares. Hmmmm....Difference and sum of 2 squares?

      Ok x = a^2-b^2 for some a and b. Now notice that when

      n is 1 x = 3 = 4 - 1 and z = 5 = 4+1

      n is 2 x = 5 = 9 - 4 and z = 13 = 9+4

      n is 3 x = 7 = 16 -9 and z = 25 = 16+9

      ...

      Sure Looks like a pattern to me. Yep, kooks like x = a^2-b^2 and z = a^2 + b^2.

      We are almost there. We now need to find y. Lets go ahead and assume this for

      x and z.



      y^2 = z^2 - x^2 = (a^2+b^2) - (a^2-b^2) =

      (a^4+2a^2*b^2+b^4) - (a^4-2a^2*b^2+b^4) = 4a^2*b^2

      so y = 2ab and the picture show now is

      if x^2+y^2=z^2 then

      x = a^2-b^2

      y = 2ab

      z = a^2+b^2



      Lets try it out. Let a =7,b=2 then

      45^2 + 28^2 = 53^2

      It works!



      So multiply it out for the general case.



      (a^2-b^2)^2 + (2ab) = (a^2+b^2)^2



      (You can do this)


      Now we have proved the general case for all a>0,b.



      Heuristics Rock,

      El cino






      To: primenumbers@yahoogroups.com
      From: plesala@...
      Date: Tue, 14 Jul 2009 06:09:33 +0200
      Subject: [PrimeNumbers] Pythagoras theorem







      Hi,

      I figured out that there is a formula for sets of integers for which Pythagoras theorem is true.

      3 4 5

      5 12 13

      7 24 25

      The formula is

      2n + 1 2n(n + 1) 2n(n + 1) + 1

      The question I am having difficulty with is whether or not there are sets of integers that do not belong to the set which is defined by the above relation; i.e are there some integers a, b and c for which

      a^2 + b^2 = c^2 that are connected by the relationship that if a is defined as

      a = 2n + 1, then b and c have to be

      b = 2n(n + 1) and c = 2n( n + 1) + 1, respectively?

      Thank you.

      Peter.

      [Non-text portions of this message have been removed]










      [Non-text portions of this message have been removed]
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