The Diophantine equation $ax^2+bxy+cy^2=0$ has integer solutions

Question

Let $q(x,y)=ax^2+bxy+cy^2$ is a binary quadratic form in integers $x,y$. Then my question is: Find conditions on $a,b,c$ such that the Diophantine equation: $ax^2+bxy+cy^2=0$ has integer solutions $(x,y)$ with $x>0,y>0$ and $x,y$ have a common prime divisor.

Answer 1

The condition that $x,y$ have a common prime divisor does not change the question, as if $(x_1, y_1)$ is a solution to $ax^2+bxy+cy^2 = 0$ then we can multiply both sides of the equation by $p^2$ for any prime $p$ and obtain $$p^2(ax_1^2+bx_1y_1+cy_1^2)=p^2\cdot 0$$ yielding the solution $$a(px_1)^2+b(px_1)(py_1)+c(py_1)^2=0$$

So we can reduce your question to the case where $x,y$ are coprime, as there are coprime solutions if and only if there are non coprime solutions.

we can now attempt to solve this by simply treating it as if it were a quadratic equation in the variable $x$. Applying the quadratic formula gives us $$x=\frac{-by\pm\sqrt{b^2y^2-4acy^2}}{2a}$$ very nicely y factors out completely, giving us $$x=y\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

we now want to know when will this have a solution where $x$ and $y$ are both positive integers. But it isn't too difficult to see that we will get integer solutions (other than $(0,0)$ ) if and only if $\sqrt{b^2-4ac}$ is rational. and we will get positive integer solutions if also $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ is positive. for which we only need to check that $b>\sqrt{b^2-4ac}$ assuming $a,b$ and $c$ are real.

Answer 2

Express $x$:

$$x = \frac{-b y \pm \sqrt{y^2 b^2 - 4 a c y^2}}{2a} = \frac{(-b \pm d)y}{2a}$$ where $d = \sqrt{b^2 - 4ac}$.

If $d$ is irational, clearly, no solutions exist. Otherwise, we can prove that (assuming $a$, $b$ and $c$ are integers), $d$ is an integer.

Thus, a pair $y = 2a$ and $x = -b + d$ is a solution. If $\gcd({x, y}) = 1$, multiply both of them with your favourite prime number.