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Pythagorean Triples Solution

The square of the odd number

    2u+i     is    4u^2+4u+1  

Hence the sum of two odd squares is divisible by 2 but not by 4; and therefore the sum of two odd squares cannot be a square. Hence of the numbers x, y one is even. If we suppose that y is even, then x and z are both odd.

    x^2 + y^2 = z^2    

     y^2 = z^2 - x^2  

Both x and z terms in the parenthesis below will be odd and thus the sum and difference will be even

     y^2 = ( z - x )( z + x ) 

Every common divisor of ( z + x )   and     ( z - x )   is a divisor of their difference 2x. Thence, since z and x are relatively prime odd numbers we conclude that 2 is the greatest common divisor of ( z + x )  and   ( z - x ) . Then we see that each of these numbers must be twice a square, so that we may write

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

Solving for x, y, z yields

       y= 2ab            y=a^2 - b^2          z = a^2 - b^2      

Using values of a & b you can generate all the integer solutions.

A variant on the above can be done to illustrate the solution technique more

    x^2 + gy^2 = z^2    

      gy^2 = z^2 - x^2  

      gy^2 = ( z - x )( z + x )    Using the same argument as above but one of the factors on the right hand side has to account for the g.  Since the g can be in either factor there are two solutions that make up the overall solution

      gy^2 = g( 2b^2 )( 2a^2)     Matching the previous phrase with this phrase to equate right hand factor terms

  ( z + x ) =2a^2      ( z - x )=2gb^2      —-OR—-       ( z + x ) =2ga^2      ( z - x )=2b^2      

     z = a^2 + gb^2      x = a^2 - gb^2        —-OR—-      z  = ga^2 + b^2      x = ga^2 - b^2   

Example Solutions with g=2 a=4 b=1
x y z Equation Set
2 4 6  z = a^2 + gb^2  x = a^2 - gb^2  
7 4 9  z = a^2 + gb^2  x = a^2 - gb^2  







Categories: Math

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