I learned this fact and it blew my mind: given an equation
$y^{2}=x^{3}+ax+b$
and two rational solutions:
$(x_1, y_1), (x_2, y_2)$
with $x_1, y_1, x_2, y_2 \in \mathbb{Q}$, then any other solution colinear with the first two solutions is also rational; that is to say, any solution to the system:
$y^{2}=x^{3}+ax+b \\ y=y_1+\frac{y_2-y_1}{x_2-x_1}(x-x_1)$
has the property:
$x, y \in \mathbb{Q}$
How would you go about proving this? Could you perhaps sketch out the main steps of the proof?
Also, if you could, please share any resources / tutorials / textbooks that could help me learn the necessary mathematics for this proof; I've already looked at a few algebraic geometry textbooks and I found it quite difficult; my background is statistics and machine learning (PhD), but I find reading pure algebra just incredibly difficult due to the sheer number of definitions they can have on a single page and still use a large number of terms I don't know.
You don't need any advanced math for this. What you are saying is that if a degree 3 polynomial $f=ax^3+bx^2+cx+d\in \mathbb Q[x]$ has 2 roots in $\mathbb Q$, then all its roots are in $\mathbb Q$. In fact, you can alway write $f=a(x-x_1)(x-x_2)(x-x_3)$, where $a\in \mathbb Q^*$ and $x_1,x_2,x_3$ are its roots. Now assume wlog $x_1,x_2\in \mathbb Q$. Then $-a(x_1+x_2+x_3)=b\in\mathbb Q$. But $a,x_1,x_2\in \mathbb Q$ and therefore $x_3\in \mathbb Q$ as well.
If you're interested in the topic and you want a soft introduction, I recommend Silverman-Tate's book "Rational points on elliptic curves".