Distinguish which second degree diophantique equation has a solution without solving them

Question

Given the three diophantine equations:

1. $-319y^2+600yx+513y+82x=16917833643583704005951315312584860330200756832904229873976761050890255147321698862822226118724$

2. $-79y^2+600yx+379y+442x=16917833643583704005951315312584860330200756832904229873976761050890255147321698862822226118478$

3. $123y^2+1800yx+849y+966x=50753500930751112017853945937754580990602270498712689621930283152670765441965096588466678355980$

By hypothesis, I know that one has a solution and the other two do not. Using transformations, I have converted them into Legendre equations ->

I homogenize with the shape ax² + bxy + cy² + dxz + eyz + fz² = 0

1. $-319y^2+600yx+513yz+82xz- 16917833643583704005951315312584860330200756832904229873976761050890255147321698862822226118724z^2=0$

2. $-79y^2+600yx+379yz+442xz-16917833643583704005951315312584860330200756832904229873976761050890255147321698862822226118478z^2=0$

3. $123y^2+1800yx+849yz+966xz-50753500930751112017853945937754580990602270498712689621930283152670765441965096588466678355980z^2=0$

and I make the change

B = $-(b^2 - 4ac)$

A = $-(bd - 2ae)^2 + (b^2 - 4ac)(d^2- 4af)$,

to convert them in the form x ^ 2 + By ² + Az ² = 0, resulting in:

1. $x^2-600^2y^2+0z^2=0$

2. $x^2-600^2y^2+0z^2=0$

3. $x^2-1800^2y^2+0z^2=0$

Doing $y = 600y$ in the first and second case $ y = 1800y$ in the third case we obtain the equation $x^2-y^2 = 0$ in all three cases and there is no way to distinguish which has solutions and which do not. With the Legendre transformation, I want to deduce which equation has a solution, although I have come to know that the 3 equations are the same.

My questions are:

Is there any other way to distinguish which equation has solutions and which do not without solving them and without transforming them into a Legendre equation? If yes, What is it?

In the process of transformation to the Legendre equation, is it possible to differentiate which equation has solutions and which not without solving them? How?

Thank you,

Javier

Answer 1

All three equations have solutions for either variable in terms of the other, though $y$ may be nether an integer nor real.

In the case of $$-319y^2+600yx+513y+82x-C=0$$ The solution for $y$ can be found with the quadratic equation $$y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\quad\text{where}\quad a=-319, b=(600x+513), c=(82x+C)$$ The solution for $x$ can be found with simple algebra and always has a real integer solution.

The other two have similar solutions. All three require arbitrary precision but this is available at WolframAlpha. For equation $(1)$, the solution for $x$ is here and solution for $y$ is here. I know you just wanted to know, in theory, if they were solvable–– but all first and second degree equations have solutions in terms of the other and WolframAlpha can help with that.

Also, both $x$ and $y$ can have natural number solutions if you fist solve for $x$ and then plug in natural numbers for $y$ to find the final numerical solution. Just seeing the algebraic solution for $x$ in the link above show that there are an infinite number of natural number solutions for all three equations.

Update: Here are solutions in $x$. I don't know if any yield integers.

1. $x = \frac{319 y^2 - 513 y + 16917833643583704005951315312584860330200756832904229873976761050890255147321698862822226118724}{600 y + 82}\\ \land 300 y + 41\ne0$

2. $x = \frac{79 y^2 - 379 y + 16917833643583704005951315312584860330200756832904229873976761050890255147321698862822226118478}{600 y + 442}\\ \land 300 y + 221\ne0$

3. $x = \frac{-41 y^2 - 283 y + 16917833643583704005951315312584860330200756832904229873976761050890255147321698862822226118660}{600 y + 322}\\ \land 300 y + 161\ne 0$