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I was reading this paper "The Solution of the Diophantine Equation $x^2+3y^2= z^2$". I have tried to prove it.

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Why $\frac{(z-x)}{2}=3y_1^2$ and $\frac{(z+x)}{2}=y_2^2$ ?

How about this theorem?

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Why $z-x=3y_1^2$, $z+x=y_2^2$, $x=\frac{(-3y_1^2+y_2^2)}{2}$ and $x=\frac{(3y_1^2+y_2^2)}{2}$ ?

Please help me. Thank you.

Paper >> http:// www.m-hikari.com/ija/ija-2014/ija-13-16-2014/abdelalimIJA13-16-2014.pdf

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    $$x^2+ay^2=z^2$$ $$x=p^2-as^2$$ $$y=2ps$$ $$z=p^2+as^2$$ – 2017-02-21

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For homogeneous equations we want solutions with the gcd of all variables coming out to one. There turn out to be two types of answers this way. If $z$ is even we have, with $\gcd(u,v) = 1$ and $v$ odd, $$ x = 2 u^2 + 2uv - v^2, \; \; y = 2 uv + v^2, \; \; z = 2 u^2 + 2uv + 2v^2. $$ If $y$ is even we have, with $\gcd(u,v) = 1$ and $u+v$ odd, the familiar $$ x = 3 u^2 - v^2, \; \; y = 2 uv , \; \; z = 3 u^2 +v^2. $$

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    We will do the x,y,z ? – 2017-02-22