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.

If $x,y$ is a solution, so is $px,py$. Your last requirement seems extraneous. — , Dec 20 '21 at 16:13
A necessary condition is that the discriminant be a rational perfect square. — , Dec 20 '21 at 16:16
Given that $x,y\neq0$ it is obvious that this is equivalent to asking whether a quadratic equation has a rational root. And as noted, the requirement that $x$ and $y$ have a common divisor is clearly extraneous. All in all, this shows absolutely no effort at all. I hope you refrain from asking such questions in the future, and instead give them some thought yourself. — Servaes, Dec 20 '21 at 18:01
Try something. Try small examples. Try a simpler version of the problem. Give it some thought. Etc. — Servaes, Dec 20 '21 at 18:12
@Guste: It is known that there are coprime solutions if and only if there are non coprime solutions. I am asking if this result is still applicable to all homogeneous forms of deree $n$. — Safwane, Dec 21 '21 at 15:58

score 0 · Answer 1 · answered Dec 20 '21 at 16:23

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.

score 0 · Accepted Answer · answered Dec 20 '21 at 16:26

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.

You say: 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. I am asking if this result is still applicable to all homogeneous forms of deree $n$. — Safwane, Dec 21 '21 at 15:16
Yeah it should be as you can always multiply the equation by $p^n$ and then distribute the primes amongst the terms in a similar way to how I did it. It isn't even too hard to prove in general, although perhaps beyond the scope of a comment. — Moosh, Dec 21 '21 at 16:12
I am still confused: If your approach is valid (multiplying by a power of $p$,...), then I must conclude that the solutions of the equation have a common prime divisor. — Safwane, Dec 21 '21 at 16:15
Yes, and they do. Not all solutions, but any equation (of this form) that has solutions will also have solutions that are not coprime. What is the issue? — Moosh, Dec 21 '21 at 19:27

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

2 Answers2