Diophantine Equations

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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 and . 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