I'm self-studying algebraic geometry through this book and am stuck right at the start.
The author states that to find all $X, Y, Z \in \mathbb{Z}$ such that $X^2 + Y^2 = Z^2$ one can simply use this parameterization of the circle $x^2 + y^2 = 1$:
$$x = \frac{2 \lambda}{\lambda^2 + 1}, y = \frac{\lambda^2 - 1}{\lambda^2 + 1}, \text{ where } \lambda = \frac{x}{1-y}.$$
Which, the author continues, leads directly to the fact that all solutions are given by:
$X = 2lm$, $Y = l^2 - m^2$, $Z = l^2 + m^2$, with $l, m$ coprime.
I don't follow this line of reasoning at all -- how he got the parameterization, what he means by setting the parameter $\lambda$, and how all that leads to the final assertion -- which I verified by computer is indeed the set of integer solutions.
I'm also interested in how I would do this over $\mathbb{Q}$, and for arbitrary curves. The first problem in section 1 is related to proving something similar for a circle of radius 5 over the rationals :)
Thanks in advance!
Draw the circle and show the $x$- and $y$-axes. Look at a point $(x,y)$ on the circle. Draw the line that passes through that point and through the point $(0,1),$ which is also on the circle. The slope of that line is $-\lambda = (y-1)/x.$ Now we have a system of two equations: $$ -\lambda = \frac{y-1} x \\[10pt] x^2+ y^2 = 1 $$ The solution for $x$ and $y$ is what you wrote.
Observe that $x$ and $y$ are rational if and only if $\lambda$ is rational. Thus letting $\lambda$ run through $\mathbb Q$ gives all rational points on the circle (except $(0,1).$) (And this shows that there are infinitely many rational points on the circle and thus infinitely many primitive Pythagorean triples. I don't know a simpler way to show that.)
Letting rational $\lambda= \dfrac \ell m$ with $\ell,m\in\mathbb Z,$ we have \begin{align} x & = \frac{2\lambda} {\lambda^2 + 1} = \frac{2\ell m}{\ell^2 + m^2} \end{align} and then treat $y$ similarly.