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Let $A,B,C,D$ be integers such that $AD-BC= 1 $ and $ A+D = -1 $.

Show by elementary means that the Diophantine equation

$$\bigl[2Bx + (D-A) y\bigr] ^ 2 + 3y^2 = 4|B|$$has an integer solution (that is, a solution $(x,y)\in\mathbb Z^2$). If possible, find an explicit solution (involving $A,B,C,D$, of course).

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**Motivation:** I arrived at this equation after trying to find explicitly the matrix $g$ suggested by Will Jagy on his answer to this question of mine. Concretely, if $\gamma=\binom{A\ \ B}{C\ \ D}$, then $\gamma$ has order $3$ in $\operatorname{SL_2}(\mathbb Z)$. By indirect methods it can be shown that $\gamma$ is conjugated in $\operatorname{SL_2}(\mathbb Z)$ to one of the matrices $P$ or $P^{-1}$, being $P=\binom{\ \ \,0\quad1}{-1\ \ -1}$ (see studiosus’ answer to the same question.). Unfortunately this argument is rather sophisticated to my knowledge, and besides I think that a direct argument is possible.

Because of this I tried to find a *explicit* matrix $g=\binom{x\ \ y}{z\ \ w}\in\operatorname{SL_2}(\mathbb Z)$ such that $gP=\gamma g$ or $gP^{-1}=\gamma g$. The matricial equalities lead to a system of $4$ linear equations in the unknowns $x,y,z,w$ , which can be easily solved. Plugging these solutions $(x,y,z,w)$ (recall that we are considering the two possibilities of conjugation, to $P$ or $P^{-1}$) into the equation $xy-zw=1$ yields $Bx^2+(D-A)xy+(-C)y^2=\pm1$. Completing the square and using the equalities $AD-BC=1$ and $A+D=-1$ we obtain the required equation. I tried to solve it explicitly, with no success.

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Reduction can also be accomplished in a conjugacy class in $SL_2 \mathbb Z,$ in a manner very similar to that for quadratic forms. As matrices, there is a little difference: for quadratic forms, we start with symmetric $G$ and work with $P^T G P, $ for conjugacy classes we work with $P^{-1} \gamma P.$

**Lemma**: given real numbers with $|V| \geq |W| > 0,$ then either $|V+W| < |V|$ or $|V-W|< |V|.$

**Proof**: $$ |V+W| |V-W| = |V^2 – W^2| = |V|^2 – |W|^2 < |V|^2. $$

Now, we have your matrix $\gamma,$ with $(A-D)^2 -4 |BC| = -3,$ or

$$ 4|BC| = (A-D)^2 + 3. $$ Since $A+D = -1,$ we know that $A-D$ is also odd. If both $2|B|, 2|C| \geq |A-D|,$ then actually $2|B|, 2|C| \geq 1 + |A-D|.$ If so,

$4|BC| \geq (A-D)^2 + 2 |A-D| + 1, $ whence

$$ (A-D)^2 + 3 = 4 |BC| \geq (A-D)^2 + 2 |A-D| + 1, $$ or

$$ 3 \geq 2 |A-D| + 1 $$ and

$$ |A-D| = 1. $$ So, the assumption that both $2|B|, 2|C| \geq |A-D|,$ then $|A-D| = 1.$

The other case is either $2|B| < |A-D|$ or $2|C| < |A-D|.$ In this case, we use either of

$$

\left(

\begin{array}{cc}

1 & -n \\

0 & 1

\end{array}

\right)

\left(

\begin{array}{cc}

A & B \\

C & D

\end{array}

\right)

\left(

\begin{array}{cc}

1 & n \\

0 & 1

\end{array}

\right) =

\left(

\begin{array}{cc}

A-Cn & B+(A-D)n – C n^2 \\

C & D+Cn

\end{array}

\right)

$$

or

$$

\left(

\begin{array}{cc}

1 & 0 \\

-n & 1

\end{array}

\right)

\left(

\begin{array}{cc}

A & B \\

C & D

\end{array}

\right)

\left(

\begin{array}{cc}

1 & 0 \\

n & 1

\end{array}

\right) =

\left(

\begin{array}{cc}

A+Bn & B \\

-(Bn^2 + (A-D)n-C) & D-Bn

\end{array}

\right)

$$

with $n=\pm 1$

to strictly reduce $|A-D|,$ as we can have the new difference of diagonal terms either $|A-D-2Cn|$ or $|A-D+2Bn|.$ Keep doing these steps, eventually the difference of the diagonal terms is exactly $1$ in absolute value; so one of them is $0$ and the other is $-1,$ and the off-diagonal terms are $1,-1.$

So far, each possible integer matrix with determinant $1$ and trace $-1$ is conjugate to at least one of these four:

$$

\left(

\begin{array}{cc}

0 & 1 \\

-1 & -1

\end{array}

\right), \;

\left(

\begin{array}{cc}

-1 & 1 \\

-1 & 0

\end{array}

\right), \;

\left(

\begin{array}{cc}

0 & -1 \\

1 & -1

\end{array}

\right) , \;

\left(

\begin{array}{cc}

-1 & -1 \\

1 & 0

\end{array}

\right)

$$

Now, the first and second are conjugate to each other, and the third and fourth are conjugate to each other, because

$$

\left(

\begin{array}{cc}

0 & 1 \\

-1 & 0

\end{array}

\right)

\left(

\begin{array}{cc}

0 & -1 \\

1 & 0

\end{array}

\right) =

\left(

\begin{array}{cc}

1 & 0 \\

0 & 1

\end{array}

\right)

$$

and

$$

\left(

\begin{array}{cc}

0 & 1 \\

-1 & 0

\end{array}

\right)

\left(

\begin{array}{cc}

A & B \\

C & D

\end{array}

\right)

\left(

\begin{array}{cc}

0 & -1 \\

1 & 0

\end{array}

\right) =

\left(

\begin{array}{cc}

D & -C \\

-B & A

\end{array}

\right)

$$

As it happens, the first and the third, which are transposes, are not $SL_2 \mathbb Z$ conjugate, so we get two pairs, one pair just the transposes of the other pair.

Why not. IF

$$

\left(

\begin{array}{cc}

s & -q \\

-r & p

\end{array}

\right)

\left(

\begin{array}{cc}

0 & 1 \\

-1 & -1

\end{array}

\right)

\left(

\begin{array}{cc}

p & q \\

r & s

\end{array}

\right) =

\left(

\begin{array}{cc}

0 & -1 \\

1 & -1

\end{array}

\right)

$$

THEN

$$ s^2 + s q + q^2 = -1, $$ which is impossible in real numbers, since $ s^2 + s q + q^2$ is a positive definite quadratic form.

You have the binary quadratic form

$$ \color{red}{ f(x,y) = B x^2 + (D-A)xy – C y^2} $$ in your last paragraph. The discriminant is $$ \Delta = (D-A)^2 + 4 B C. $$

You also have $AD-BC = 1$ and $A+D = -1.$ So, $BC – AD = -1$ and

$$ A^2 + 2 AD + D^2 = 1, $$

$$ 4BC – 4AD = -4, $$

$$ A^2 – 2 AD + D^2 +4BC = -3, $$

$$ \Delta = (A – D)^2 +4BC = -3. $$

If we had $\gcd(A-D,B,C) > 1$ we would have a square factor of $\Delta,$ so that is out, the coefficients of $f(x,y)$ are relatively prime (as a triple, not necessarily in pairs). Next, $-3$ is not a square, so we cannot have $B=0$ or $C=0.$

Finally $f$ is definite. If, say, $B > 0,$ it is positive definite. With discriminant $-3,$ it is then equivalent in $SL_2 \mathbb Z$ to

$$g(u,v) = u^2 + u v + v^2$$ and both $g$ and $f$ integrally represent $1.$

If $B < 0,$ then $f$ is negative definite. With discriminant $-3,$ it is then equivalent in $SL_2 \mathbb Z$ to

$$h(u,v) = -u^2 – u v – v^2$$ and both $h$ and $f$ integrally represent $-1.$

That is enough for what you asked. I should point out that the reduction for positive binary forms is very similar to the Euclidean algorithm for finding GCD is very similar to the algorithm for finding the matrix $W$ in $SL_2 \mathbb Z$ that is conjugate to your original and has, for example, the smallest value of

$(w_{22} – w_{11})^2.$ I recommend Buell, Binary Quadratic Forms if I have not already. I was going to answer with a reduction that minimized $(A-D)^2,$ then looked again at your question and realized there was a shortcut. Note that many books do reduction for positive definite binary forms, and give short lists for discriminants of small absolute value.

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