General quadratic diophantine equation.

Question

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!

Answer 1

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 $u^2-pv^2 = \pm 1$.)

Answer 2

The method suggested by the author of the question. This method is Diofantos geometry. The downside is the need to know the first solution. Then build a secant and look for the next solution. We are always bound to the coefficients.

If you use an algebraic approach - you can get the right solution. I wrote another formula.

Though it is necessary to bring the decisions some pretty simple solutions:

the equation: $$aX^2+bXY+cY^2=f$$

If the root of the whole: $\sqrt{\frac{f}{a+b+c}}$

Then use the solution of Pell's equation: $p^2-(b^2-4ac)s^2=1$

Solutions can be written: $$Y=((4a+2b)ps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a+b+c}}$$ $$X=(-(4c+2b)ps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a+b+c}}$$

If a root: $\sqrt{fa}$ then the solutions are of the form: $$Y=4ps\sqrt{fa}$$ $$X=(-2bps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a}}$$
Although it should be mentioned, and the equation: $aX^2-qY^2=f$

If the root of the whole: $\sqrt{\frac{f}{a-q}}$ Using equation Pell: $p^2-aqs^2=1$

solutions can be written: $$Y=(2aps\pm(p^2+aqs^2))\sqrt{\frac{f}{a-q}}$$ $$X=(2qps\pm(p^2+aqs^2))\sqrt{\frac{f}{a-q}}$$

And for that decision have to find double formula. $$Y_2=Y+2as(qsY-pX)$$ $$X_2=X+2p(qsY-pX)$$

Be aware that this formula in General, therefore, should be considered equivalent to a quadratic form. That is, for example, to do this change $X\longrightarrow{(X+kY)}$

For some special cases, you can record and more than a simple formula.

For a private quadratic form: $Y^2=aX^2+bX+1$

Using solutions of Pell's equation: $p^2-as^2=1$

Solutions can be expressed through them is quite simple. $$Y=p^2+bps+as^2$$ $$X=2ps+bs^2$$ $p,s$ - can be any character.

Solutions of the equation: $ax^2-by^2+cx-dy+q=0$

you can record if the root of the whole: $k=\sqrt{(c-d)^2-4q(a-b)}$

Then using the solutions of the equation Pell: $p^2-abs^2=\pm1$

Then the formula of the solution, you can write: $$x=\frac{\pm1}{2(a-b)}(((d-c)\pm{k})p^2+2(bk\mp(bc-ad))ps+b(a(d+c)-2bc\pm{ak})s^2)$$ $$y=\frac{\pm1}{2(a-b)}(((d-c)\pm{k})p^2+2(ak\mp(bc-ad))ps-a(b(d+c)-2ad\mp{bk})s^2)$$

If the root is a need to find out if this is equivalent to the quadratic form in which the root of the whole. This is usually accomplished this replacement: $x$ in such number $x+ty$ Forgot to say. The characters inside the brackets do not depend on the sign of the Pell equation.

It depends only before $\pm{1}$

The algebraic approach is better because it allows us to answer such questions for which Diophantos geometry in General meaningless. For example why curves for triangular numbers, solving equations in integers is always there. If any coefficients.

The solution there. Curves triangular numbers.

To convince supporters of Diophantos geometry is not possible. Although I think a purely algebraic approach allows us to write the formula for the solutions of the equation. And in this case it all comes down to solving the Pell equation. And this is a standard task and it normally is solved.