Here is my problem: I am given a general quadratic diophantine equation:
$$ax^2 + bxy + cy^2 + dx + ey + f = 0$$
where $x$ and $y$ are variables with integers $a,b,c,d,e,f$. I have to show that if the equation has one solution in the set of rational numbers, the equation will have an infinite amount of solutions in the set of rational numbers. I am given three steps to help me prove this:
a) Name the rational solution $(x_0,y_0)$. Write the equation of a straight line $L$ through $(_0,y_0)$ with slope $t$, where $t$ is a rational number.
b) Show that the second point of intersection $(x_1,y_1)$ with line $L$ is a rational solution as well. Show this without calculating the second point of intersection explicitly by using the following theorem, which you also have to prove:
If $ax^2 + bx + c = 0$ is an equation with $0\neq a, b, c\in\mathbb Q$ and solutions $x_0$ and $x_1$, then $x_0\cdot x_1 = c/a$.
c) You can now show that original diophantine equation has an infinite amount of rational solutions.
If there are any ambiguities in my question, I will be happy to try and make them more understandable as english is not my mother languange.
Thank you in advance!
Best Answer
This is an old post but may be useful to those who come across it. The problem can be reduced to an identity. Given,
$$ax^2+bxy+cy^2+dx+ey+f=0\tag{1}$$
for integer constants $a,b,c,d,e,f$. First, solve it as an eqn in $y$,
$$y = \frac{-e-bx \pm \sqrt{ px^2+qx+r}}{2c}\tag{2}$$
where,
$$p,\;q,\;r = (b^2-4ac),\; -2(2cd-be),\;(e^2-4cf)\tag{3}$$
Thus, if the discriminant of $(2)$ is a square, or you have an initial rational solution to,
$$px_0^2+qx_0+r = t^2\tag{4}$$
then it implies $(2)$ is rational.
So here is the relevant identity. Let $p,q,r$ be defined as above. Then,
$$\begin{aligned} &ax^2+bxy+cy^2+dx+ey+f=\frac{(px_0^2+qx_0+r)-t^2}{-4c}=0\\ \text{where},\qquad\\ &y = \frac{-e-bx \pm \big(u/v(x-x_0)+t\big)}{2c}\\ &x = x_0+\frac{-2tuv+(2px_0+q)v^2}{u^2-pv^2} \end{aligned}\tag{5}$$
for arbitrary $u,v$. So if you have initial rational solution $x_0$ to $(4)$, then the identity $(5)$ shows you can generate an infinite more.
(P.S. Furthermore, if $c=1$, and non-square $p=b^2-4ac>0$, then you can find integer $x,y$ by solving the Pell equation $u^2-pv^2 = \pm 1$.)