On a Quick Expansion Method for Pythagorus Triples

by Frank C. Fung ( 1st published in December, 2021. )

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Section IX - Setting up the { Value-In-Question } - [ V-sub-IQ ]

Summary for the Section :

• In this Section , we shall give full details as to :
  • how we happen to stumble upon the Quick Expansion Method for Pythagorus Triples .

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Recalling the Author's previous paper :

• Let us recall that :
  • in the author's previous paper :
    • On Breaking-down any Triangle into a pair of mirror-imaged Equilateral Triangles

    our first aim there was to solve for the 2 Equilateral Triangles for any given Triangle .

  It was therefore appropriate for us to do an illustration of the 3 Triangles , namely :

  • the given Triangle { Triangle A-B-C } itself ,

    and

  • the 2 mirror-imaged Equilateral Triangles { Triangle T1-T2-T3 } and { Triangle U1-U2-U3 } respectively ;

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Setting up the new { Triangle A-B-C } :

• And one of the few triangles we did construct for the above-said illustration purposes was this new { Triangle A-B-C } as shown below ,
  • with :
    • 
    • 
    • 

  

  And we take special note here of the construction procedure for this type of { Triangle A-B-C } :

  • We first bring in 2 { Pythagorus-Triple Triangles } of the same height :
    • the 1st triangle being the { 20-15-25 } Triangle based on the { 4-3-5 Pythagorus Triple } ,
      • as shown in yellow-color in the diagram in-the-middle above ;

    • the 2nd triangle being the { 36-15-39 } Triangle based on the { 12-5-13 Pythagorus Triple } ,
      • as shown in green-color in the diagram on-the-right above .

    And the above-said 2 triangles would then define and confirm the positions of [ Point B ] and [ Point C ] respectively on the Complex Plane .

  • And [ Point A ] would always lie on the the [ REAL Axis ] and length of [ Vector O-A ] would be given by :
    • the sum of widths of the 2 above-said triangles , and in this case :
      • { Length of [ Vector O-A ] } = 20 + 36 = 56
        • as shown in the diagram on-the-left above .

• As such :
  • the Point-Of-Origin [ Point O ] is always the centroid of { Triangle A-B-C } , arising from :
    • 

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Establishing the numeric values for [ K2 ] / [ K1 ] / [ K0 ] :

• Based on the values of [ A ] / [ B ] / [ C ] being :
  • 
  • 
  • 

  we then have :

  • First :
    • 

      thereby confirming that [ Point O ] the Point-Of-Origin is in fact the centroid of { Triangle A-B-C } ;

  • Secondly :
    • 
    • 
    • 

    Consequently :

    • 

  • Third-and-finally :
    • 

• Let us recall that :
  • In order to solve for the 2 mirror-imaged Equilateral Triangles associated with the given { Triangle A-B-C } ,
    • we shall assume that the complex values [ A ] / [ B ] / [ C ] are the solution values
      • for a { 3rd-Order Polynomial Equation } of the format :
        • 

      so that :

      • these 2 equations below shall be considered equivalent throughout this Section here .
        • 

        • 

  As such :

  • upon expansion of the 2nd equation above , we have :
    • 

    • 

    • 

• Let us recall that we have established above :
  • 

  • 

  • 

  Therefore :

  • 

  • 

  • 

  Consequently we can now re-write the original { 3rd-Order Polynomial Equation } ,

  • i.e. this equation :
    • 

    in this revised format :

    • 

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Establishing the complex value [ V-sub-IQ ] :

• Since the revised equation is now :
  • 

  we can then follow the solution procedure established in the last 3 Sections to arrive at :

  • 

  • 

    in order to further solve for the values of [ T ] via :

    • either :
      • 

      or :

      • 

• Let us now concentrate on the { quantity under the square-root sign } in both [ W1 ] and [ W2 ] .

  And we take note here that :

  • 

  • 

  Consequently , we can re-express the { quantity under the square-root sign } in this manner :

  • 

    yielding :

  • 

  And on further manipulations , we have :

  • 

  And finally :

  • 

• We then set up the complex value [ V-sub-IQ ] , being the { Value-In-Question } :
  • 

  And this definition of the [ V-sub-IQ ] shall be used throughout the remainder of the paper .

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Doing the calculations for [ V-sub-IQ ] :

• Let us now do the calculations for [ V-sub-IQ ] .

  And we bring in from above :

  • for [ K0 ] :
    • 

      yielding :

    • 

      and further :

    • 

  • for [ K1 ] :
    • 

      yielding :

    • 

      and further :

    • 

• And based on the definition of [ V-sub-IQ ] established above :
  • 

  we then have this numeric value for [ V-sub-IQ ] :

  • 

• And we happen to take note at this point that :
  • the { REAL part } and the { IMAGINARY part } of [ V-sub-IQ ] do form a Pythagorus Triple :

  	Value of Number	Square of Number	Factors of Number
  X	7,545,027,329	56,927,437,395,356,874,241	17 x 17 x 7 x 1409 x 2647
  Y	13,044,327,120	170,154,470,013,567,494,400	17 x 17 x 16 x 9 x 5 x 11 x 41 x 139
  xxxxx	xxxxxxxxxxxxxxxxxxxxxxxxx	xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx	xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
  Z	15,069,237,121	227,081,907,408,924,368,641	17 x 17 x 17 x 353 x 8689

  And that is to say :

  • For :
    • X = 7,545,027,329

    • Y = 13,044,327,120

    • Z = 15,069,237,121

    then :

    • { X / Y / Z } is a Pythagorus Triple .

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What's Next :

• This is rather interesting and we shall look at more data in Part IV , next .

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go to the next Section : Section X - Outline and Purpose for Part IV

go to the last Section : Section VIII - Solving the 3rd-Order Polynomial Equation from { Triangle A-B-C }

return to the Pythagorus Triples Homepage

Original dated 2021-12-27