Second Degree Diophantine Equation, When exists solutions?

Question

In second degree diophantine equations, ax²+bxy+cy²+dx+ey+f=0,

Is there a theorem or criterion that allows us to decide whether or not any second degree diophantine equation has solutions without resolution?

Thank you,

Javier

Answer 1

There is a procedure that can be learned by a beginner. It is sufficiently messy that I won't give the giant expressions when attempting to finish using symbols.

First, make a new variable $z$ and make the polynomial homogeneous. If there are integers $x,y,z$ that make the expression below zero, then there are rational solutions to your original problem

$$ a x^2 + bxy + c y^2 + dzx + e yz + f z^2 $$

Next, the Hessian matrix of second partial derivatives is

$$ H= \left( \begin{array}{ccc} 2a&b&d \\ b&2c&e \\ d&e&2f \end{array} \right) $$

Next, there is a procedure for constructing $P^T H P = D,$ where $P$ is rational determinant $\pm 1$ and $D$ is rational diagonal.

Here is an example I made up.

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 3 }{ 2 } & 1 & 0 \\ - 9 & \frac{ 11 }{ 3 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 3 & 7 \\ 3 & 6 & 5 \\ 7 & 5 & 10 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 3 }{ 2 } & - 9 \\ 0 & 1 & \frac{ 11 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 3 }{ 2 } & 0 \\ 0 & 0 & - \frac{ 104 }{ 3 } \\ \end{array} \right) $$

I deliberately adjust $P$ to $PW,$ where I chose $W$ to be the diagonal matrix $1,2,3.$ The new version is

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ -3 & 2 & 0 \\ - 27 & 11 & 3 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 3 & 7 \\ 3 & 6 & 5 \\ 7 & 5 & 10 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 3 & - 27 \\ 0 & 2 & 11 \\ 0 & 0 & 3 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & - 312 \\ \end{array} \right) $$

Next are there any rational solutions to $$ 2 u^2 + 6 v^2 - 312 w^2 = 0, $$ or $$ u^2 + 3 v^2 - 156 w^2 = 0. $$ If there are rational solutions we get integer solutions by multiplying by a common denominator.

This is the province of Legendre's theorem on indefinite ternaries.

We note $$ 156 = 4 \cdot 3 \cdot 13. $$ $ u^2 + 3 v^2 - 156 w^2 = 0. $ We can express $12 = 9 + 3\cdot 1$ and $13 =1 + 3 \cdot 4 $

The standard way to combine these expressions gives two versions, $$ 156 = 9^2 + 3 \cdot 5^2 = 3^2 + 3 \cdot 7^2 $$

We have constructed two solutions to our intermediate problem $ u^2 + 3 v^2 - 156 w^2 = 0. $ One is $u=9, \; v = 5, \; w = 1.$ Another is $u=3, \; v = 7, \; w = 1.$

It happens that there is a way to parametrize all integer triples that satisfy $ x^2 + 3 y^2 + 5 z^2 + 5yz+7zx+3xy=0 $ The idea goes back to Fricke and Klein (1897). As we now know there are integer solutions to

$$x^2 + 3 y^2 + 5 z^2 + 5yz+7zx+3xy=0$$

3 1 3 transposed 3 1 -1 1 -5 -1 transposed $$ x = 3 p^2 + pq + 3 q^2$$ $$ y = p^2 - 5pq - q^2 $$ $$ z = -p^2 +3pq -3 q^2 $$ 1 -5 3 -1 3 -3 transposed 3 -1 -3

tells us infinitely many integer solutions $$ x = 3 p^2 + pq + 3 q^2$$ $$ y = p^2 - 5pq - q^2 $$ $$ z = -p^2 +3pq -3 q^2 $$ Note how very much this resembles the parameterization of Pythagorean Triples.

A little more work would provide parameterizations of all solutions.

-3 1 -3 transposed -3 -1 1 -1 -5 1 transposed 1 -5 3 1 3 3 transposed -3 1 3

is better, as it tells us directly when we can have $z=+1.$ Only finitely many times as the last binary is positive definite.
$$ x = -3 p^2 + pq - 3 q^2$$ $$ y = -p^2 - 5pq + q^2 $$ $$ z = p^2 +3pq +3 q^2 $$

Yep, the form for $z$ keeps coming up discriminant $-3,$ therefore positive. Still, we then have a few solutions to $x^2 + 3 y^2 + 5 + 5y+7x+3xy=0$

Answer 2

There is a whole theory about this. It's beautiful too. See these books for instance:

• Binary Quadratic Forms: An Algorithmic Approach by Bachmann and Vollmer

• Rational Quadratic Forms by Cassels

Answer 3

All second degree equations are solvable for one of the variables using the quadratic equation.

\begin{equation} ax^2+bxy+cy^2+dx+ey+f=0 \end{equation} \begin{equation} \implies x = -\frac{\pm\sqrt{(b y + d)^2 - 4 a (y (c y + e) + f)} + b y + d}{2 a} \quad \text{ for }\quad a\ne0\\ \text{or}\quad x = -\frac{y (c y + e) + f}{b y + d}\quad\text{ for }\quad a = 0 \land b y + d\ne0 \end{equation}

\begin{equation} \implies y = -\frac{\pm\sqrt{(b x + e)^2 - 4 c (x (a x + d) + f)} + b x + e}{2 c} \quad\text{for}\quad c\ne0\\ \text{or}\quad y = -\frac{x (a x + d) + f}{b x + e} \quad \text{for}\quad c = 0 \land b x + e\ne0 \end{equation}

As long as the discriminant (the part under the radical) is non-negative, there are unlimited solutions.