Answer to Question #200851 in Combinatorics | Number Theory for Sasah Solomon

Question

Proof of rational points on the general form of conic sections?

Answer 1

 There is a theorem that states :

if a conic has 1 rational solution it has infinitely many rational solutions.

WE CAN PROVE IT AS FOLLOWS:

Suppose you have a rational conic as:

$ax^2+bxy+cy^2+dx+ey+f=0.$

A generic line intersects a conic in 2 points.

If the conic and the line are rational {given by rational coefficients},

then the set consisting of those two points is invariant by Q-automorphisms,

so either each point is rational, either they are defined over a quadratic extension of Q

Q and the automorphism of that extension swaps the two intersection points.

If you have a rational point P on the conic,

then the rational lines going through it will intersect the rational conic at another rational point (possibly P itself if the line is tangent at P to the conic)

Moreover, two different lines going through P can only intersect at P

, so the lines that are not tangent to the conic at P

give you distinct points on the conic.

This actually gives you a bi-rational map between P1.

and the conic.

proof:

Suppose we have x,y $\epsilon$ Q such that g(x, y) = 0. Consider the line through (x, y) with rational direction vector ( u ) parameterized by ( x,y )+t( u ). We claim that the second intersection point of the line and the conic is also a rational point (the first intersection point is (x, y), corresponding to t= 0.

g(x+tu,y+t) = a(x+tu)²+b(x+tu)(y+t)+c(y+t)²+

d(y+tu)+e(y+t) +f^g(x,y) =0

= t²(au²+bu+c)+t(du+e+2aux+bx+byu+2y)

So the second intersection point corresponds (consider g(x+ tu, y+ t) = 0) to the rational parameter( t).

There are clearly infinitely many ways to choose u $\epsilon$ Q such that t represents a nontrivial rational number, giving rise to infinitely many rational points on the curve of the form g(x+tu,y+t).