# Polynomial of degree two in two variables and rational points

This question of mine and the comments on the answer led me to the following more general problem:

Suppose that we have polynomial $f(x,y)=a_1xy+a_2x^2+a_3y^2+a_4x+a_5y+a_6$:

Consider now the equation $f(x,y)=0$:

I would like to know the following:

What are the most general conditions that you know of on the coefficients $a_1.a_2.a_3,a_4,a_5,a_6$ of the polynomial so that under those conditions there are infinitely many points $(x,y)$ with both rational coordinates such that $f(x,y)=0$?

• There are either no rational points or infinitely many. Proof. – vadim123 Nov 13 '15 at 18:01
• @vadim123 Is that the case when all the coefficients are rational? – Farewell Nov 13 '15 at 18:06
• The equation $x^2+y^2+1=0$ has no rational solutions. – vadim123 Nov 13 '15 at 18:18
• @vadim123 Okay, so if all the coefficients are rational that does not mean that there will be infinitely many rational points, but the question is about are there some general conditions that you know of under which there will be infinitely many rational points? – Farewell Nov 13 '15 at 18:22
• If we look for solutions in rational numbers you can use these formulas. math.stackexchange.com/questions/1513733/… If you are looking for in integers, it must be reduced to a Pell equation. For example in this case. artofproblemsolving.com/community/c3046h1048219 – individ Nov 14 '15 at 5:45

I am assuming you are working over a number field $K$. Then, $f(x,y)$ describes a conic section. If you homogenize it by adding suitable powers of $z$ to every monomial, you get a ternary quadratic form, which can always be diagonalized by suitable change of variables to obtain something of the form $Q(x,y,z) = ax^2 + by^2 + cz^2$. You are then asking when the quadratic form is isotropic (ie, there is a triple $(x,y,z)\ne (0,0,0)$ such that $ax^2 + by^2 + cz^2 = 0$).
Since the Hasse Principle holds for quadratic forms, $Q = 0$ has a solution over a number field $K$ if and only it has solutions over every $K_\nu$, where $\nu$ is a place of $K$ and $K_\nu$ is the completion of $K$ at $\nu$. If $K = \mathbb{Q}$, then $\{K_\nu\}_\nu = \{\mathbb{Q}_p\}_{p\text{ prime}}\cup\{\mathbb{R}\}$. (Here $\mathbb{Q}_p$ is the field of $p$-adic numbers).
Over $\mathbb{R}$, $Q = 0$ has solutions if and only if $a,b,c$ are not all positive and not all negative.
At any finite place $\nu$, you can scale $Q$ by some number so that $a,b,c$ are all $\nu$-adically integral, and one of them is $-1$. WLOG suppose $c = -1$. In this case, $Q = 0$ has a solution iff the Hilbert symbol $(a,b)_p = 1$. If $K = \mathbb{Q}$ you can find formulas for it here: https://en.wikipedia.org/wiki/Hilbert_symbol