In Weissman [1], there is a nice description of how to generate a mapping between reduced fractions and pythagorean triples. In outline the method is as follows:
Take the unit circle $x^2+y^2=1$ and a straight line through $(0,1)$ with a rational gradient $m$, i.e. $y = mx + 1$.
The line and the circle intersect at $(0,1)$ and $(u,v)$. The coordinates of the second point can be found to be
$$ u = \frac{- 2 m}{{{m}^{2}}+1} \qquad \text{ and } \qquad v = \frac{{ - {m}^{2}} + 1}{{{m}^{2}}+1} $$As $m$ is rational, we can write it as $a/b$ with gcd$(a,b)=1$ and we can re-write $u$ and $v$. $$ u = -\frac{2 a b}{{{b}^{2}}+{{a}^{2}}} \qquad \text{ and } \qquad v = \frac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}+{{a}^{2}}} $$
Using the fact that $u^2+v^2 = 1$ leads to $$ (2ab)^2 + (b^2-a^2)^2 = (b^2 + a^2)^2 $$ which gives a pythagorean triple $2ab$, $b^2-a^2$ and $a^2+b^2$. It is possible to show that only 1 is a common divisor for the triple.
Subsequently, there is an exercise to show that there are an infinite number of a variant of pythagorean triples $(x,y,z)$, with the form $x^2 + 2 y^2 = 3 z^2$.
The hint given is to use the point $(1,1)$ instead of the point $(0,1)$ for the line in a similar argument to that given above, i.e. to use the line $y = m(x-1) + 1$.
When I do this, still using the unit circle, I get quite complex expressions for the intersection coordinates. For example, one intersection point has $$ u = \frac{{{m}^{2}} - m - \sqrt{2m}}{{{m}^{2}}+1} \qquad \text{ and } \qquad v = \frac{{-{m}^{2}}-2m+1}{{{m}^{2}}+1} $$ Substituting in an expression for a reduced fraction makes things even more complicated and I cannot link the expressions for $u$ and $v$ to a triple of the form described.
Any advice on how I might proceed would be gratefully received.
[1]: Weissman, Illustrated Theory of Numbers (AMS)
Follow up.
After a nudge from Lord Shark (use the ellipse $x^2 + 2 y^2 = 3$ instead of the unit circle), expressions for $(x,y,z)$ satisfying $x^2 + 2 y^2 = 3 z^2$ can be found. I now make these to be: $$ x = 2{{a}^{2}} -4ab -{{b}^{2}} \quad y = {{b}^{2}}-2ab-2{{a}^{2}} \quad z = {{b}^{2}}+2{{a}^{2}} $$
The point $(1,1)$ is not on the unit circle. It is on the ellipse $E$ with equation $$x^2+2y^2=3$$ which is the relevant curve in this context. The second intersection of the line through $(1,1)$ of slope $m$ with $E$ can be written $(f(m)/h(m),g(m)/h(m))$ where $f$, $g$ and $h$ are simple polynomials with integer coefficients. Then $f(m)^2+2g(m)^2=3h(m)^2$ etc.