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This question is a continuation of the following two questions:

Discriminants of indefinite integral binary quadratic forms admitting 3 or 6 torsion.

On certain solutions of a quadratic form equation

Let $f$ be a binary quadratic form with integer coefficients and non-zero discriminant. For $T = \begin{pmatrix}t_1 & t_2 \\ t_3 & t_4 \end{pmatrix} \in \operatorname{GL}_2(\mathbb{R})$, define $f_T(x,y) = f(t_1 x + t_2 y, t_3 x + t_4 y)$. Put

$$\displaystyle \operatorname{Aut}(f) = \{T \in \operatorname{GL}_2(\mathbb{Z}) : f_T = f\}.$$

When $f$ is positive definite, then $\# \operatorname{Aut}(f)$ is easy to determine. In particular, if $f(x,y)$ is reduced, so that it is written as

$$\displaystyle f(x,y) = ax^2 + bxy + cy^2$$

with $|b| \leq a \leq c$ with $ac > 0$, then $\# \operatorname{Aut}(f) = 12$ when $a = b = c = 1$, $\# \operatorname{Aut}(f) = 8$ when $a = c = 1, b = 0$, $\# \operatorname{Aut}(f) = 4$ when $a = \pm b$, and $\#\operatorname{Aut}(f) = 2$ otherwise. When $f$ is reducible, the situation is also easy to account for. The most difficult case is when $f$ is indefinite and irreducible.

One part of the problem is easy. If we consider only positive determinant elements, then they are given by

$$\displaystyle \begin{pmatrix} \dfrac{s+bt}{2} & & -ct \\ \\ at & & \dfrac{s - bt}{2} \end{pmatrix},$$

where $(s,t)$ is a solution to the Pell equation $x^2 - (b^2 - 4ac)y^2 = 4$ (note that this equation is always soluble since $(2,0)$ is a solution).

The problem is the elements of negative determinant. I can show (see the second question linked above) that such elements take the form

$$\displaystyle \begin{pmatrix} m & (bm + cn)/a \\ n & - m\end{pmatrix},$$

where $(m,n)$ is a solution to $am^2 + bmn + cn^2 = a$, which is again always soluble. But there is no guarantee that $(bm+cn)/a$ is an integer. This is guaranteed if $n$ is co-prime to $a$, since then looking at the equation above we see that $a | n(bm+cn)$, whence $a | bm + cn$. There are however examples where this doesn't happen; I believe $5x^2 + 3xy - 11y^2$ is such an example. All of the solutions to $5x^2 + 3xy - 11y^2 = 5$ have $5 | y$, and since $5 \nmid 3$ and $\gcd(m,n) = 1$ by necessity, it is impossible for $3m-11n$ to be divisible by 5 and thus no way for the above matrix to be integral.

However, such examples seem extremely rare. I don't have a lot of computational evidence except to say that there is no such example with $D = b^2 - 4ac < 100$.

The question is, when is the set

$$\displaystyle G^-(f) = \{T \in \operatorname{Aut}(f) : \det T = -1\}$$

non-empty? There are several sufficient conditions for this to be non-empty: if $f$ is equivalent to a diagonal form, if $f$ is equivalent to a form such that $a | b$, , if $f$ is equivalent to a symmetric form, or $f$ is equivalent to a form such that there exists a solution $(m,n)$ to $am^2 + bmn + cn^2 = a$ with $\gcd(n,a) = 1$. Moreover, if the class number is one, then all forms with discriminant $d$ are equivalent and therefore must be equivalent to either $x^2 - (d/4)y^2$ if $d$ is even or $x^2 + xy - (d-1)y^2/4$ if $d$ is odd. Both cases have non-empty $G^-$.

A related question is, suppose that $f_1(x,y) = a_1 x^2 - b_1 y^2$ and $f_2(x,y) = a_2 x^2 - b_2 y^2$ are two non-proportional binary quadratic forms such that $a_1 b_1 = a_2 b_2 > 0$. Then when can $f_1, f_2$ be $\operatorname{GL}_2(\mathbb{Z})$-equivalent?

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  • $\begingroup$ the second part I can do: both your forms are "diagonal" and so "ambiguous." They are $GL_2 \mathbb Z$ equivalent if and only if they are $SL_2 \mathbb Z.$ For any specific forms, this can be checked with the Lagrange cycles of reduced forms, or more recent Zagier cycles of reduced forms. $\endgroup$ – Will Jagy Jul 27 '16 at 18:18
  • $\begingroup$ @WillJagy I am interested in doing asymptotic estimates, not algorithms in checking specific cases, so I will need some sort of theorem that can count over a long range of discriminants which forms $f$ has non-empty $G_f^-$. For example, it would be nice if the diagonal forms above are pairwise distinct for different divisors, but it seems to be not the case... $\endgroup$ – Stanley Yao Xiao Jul 27 '16 at 18:20
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    $\begingroup$ From page 144, Theorem 103, in Dickson 1929, the number of ambiguous classes is at least as large as the number of genera. This is in section 83, where he is just discussing forms with even middle coefficient. So, if there is just one genus, as in $x^2 - p y^2$ when prime $p \equiv 1 \pmod 4,$ the only ambiguous class is that of the principal form. For $x^2 - q y^2$ when $q \equiv 3 \pmod 4,$ there are just two ambiguous classes, principal and $-x^2 + q y^2.$ $\endgroup$ – Will Jagy Jul 27 '16 at 19:23
  • $\begingroup$ The formula I've shown, but it is still not satisfied. It is not clear why? For quadratic forms $ax^2+bxy+cy^2=j$ It is necessary that the ratios between them were connected through the square. Or so $t^2=ja$ Or so $t^2=j(a+bk+ck^2)$ . You can pick up a different configuration, but the task still comes down to this. $\endgroup$ – individ Jul 28 '16 at 4:39
  • $\begingroup$ There is another approach. But there is quite a long formula and You formula not like. The meaning is simple to find solutions for the equation $Z=\pm1$ math.stackexchange.com/questions/1513733/… You can use another idea. To reduce the equation to another form. $ax^2+bxy+cy^2=(p^2-ks^2)(t^2-qn^2)=j$ $\endgroup$ – individ Jul 28 '16 at 6:20
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Here is a necessary and sufficient condition for $G^{-}(f)$ to be non-empty, taken from [1, Exercise 6.21]:

Let $S = \begin{pmatrix} a & b/2 \\ b/2 & c \end{pmatrix}$ with $a,b$ and $c \in \mathbb{Z}$. Then the following are equivalent:

$(1)$ There exists $A \in \text{GL}_2(\mathbb{Z})$ with $\det(A)= -1$ such that $^tASA = S$.

$(2)$ The matrix $S$ is $\text{SL}_2(\mathbb{Z})$-equivalent to $\begin{pmatrix} a & -b/2 \\ -b/2 & c \end{pmatrix}$.

$(3)$ There exists $S' = \begin{pmatrix} a' & b'/2 \\ b'/2 & c' \end{pmatrix}$ with $a', b'$ and $c' \in \mathbb{Z}$ which is $\text{SL}_2(\mathbb{Z})$-equivalent to $S$ such that $b'$ is divisible by $a'$. (Here we may take $b' = a'$ or $b' = 0$.)

Hint: If $\,^tASA = S$ with $\det(A)= -1$, then show that $\text{Tr}(A) = 0$ and $A^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Then reduce the case when $A$ is upper triangular.

If the above conditions hold, the $\text{SL}_2(\mathbb{Z})$-class to which such an $S$ belongs is called an ambig class (sic). This is also what John Robertson calls an ambiguous class in [2]. From this source, we get

Let $D \in \mathbb{Z}$ which is not a square and such that $D \equiv 0$ or $D \equiv 1 \,(\text{mod } 4)$. Then the number of ambiguous classes is the order of the genus group of $D$, that is, the order of $H^{+}(D) \otimes \mathbb{Z}/2\mathbb{Z}$ where $H^{+}(D)$ is the strict class group of $D$.


[1] T. Ibukiyama, M. Kaneko, "Quadratic Forms and Ideal Theory of Quadratic Fields" in Bernoulli Numbers and Zeta Functions, pp 75-93, 2014.
[2] J. Robertson, "Computing in Quadratic Orders", 2009.

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This may or may not be what you want... my sense is that the conditions you describe as sufficient are also necessary, in this case the bit about equivalent to a form $\langle a,b,c \rangle$ such that $a | b.$ This is from Dickson's Introduction to the Theory of Numbers, 1929. Well, you can judge for yourself.

enter image description here

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Just illustrating Dickson's point, that ambiguous forms occur in the cycle of reduced forms of a given form

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 5 0 -17

  0  form              5           0         -17  delta      0
  1  form            -17           0           5  delta      1
  2  form              5          10         -12


          -1          -1
           0          -1

To Return  
          -1           1
           0          -1

0  form   5 10 -12   delta  -1     ambiguous  
1  form   -12 14 3   delta  5
2  form   3 16 -7   delta  -2
3  form   -7 12 7   delta  2
4  form   7 16 -3   delta  -5
5  form   -3 14 12   delta  1
6  form   12 10 -5   delta  -2
7  form   -5 10 12   delta  1     ambiguous            -1 composed with form zero  
8  form   12 14 -3   delta  -5
9  form   -3 16 7   delta  2
10  form   7 12 -7   delta  -2
11  form   -7 16 3   delta  5
12  form   3 14 -12   delta  -1
13  form   -12 10 5   delta  2
14  form   5 10 -12


  form   5 x^2  + 10 x y  -12 y^2 

minimum was   3rep   x = -1   y = -1 disc 340 dSqrt 18  M_Ratio  12.96
Automorph, written on right of Gram matrix:  
-130789  -371952
-154980  -440749
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jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$

For your example, it is not ambiguous

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 5 3 -11

  0  form              5           3         -11  delta      0
  1  form            -11          -3           5  delta      1
  2  form              5          13          -3


          -1          -1
           0          -1

To Return  
          -1           1
           0          -1

0  form   5 13 -3   delta  -4
1  form   -3 11 9   delta  1
2  form   9 7 -5   delta  -2
3  form   -5 13 3   delta  4          -1 composed with form zero  
4  form   3 11 -9   delta  -1
5  form   -9 7 5   delta  2
6  form   5 13 -3


  form   5 x^2  + 13 x y  -3 y^2 

minimum was   3rep   x = 0   y = 1 disc 229 dSqrt 15  M_Ratio  9
Automorph, written on right of Gram matrix:  
-16  -45
-75  -211
=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$
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  • $\begingroup$ Is the question you are answering that this form is not equivalent to a diagonal form? $\endgroup$ – Stanley Yao Xiao Jul 27 '16 at 18:31
  • $\begingroup$ @StanleyYaoXiao, no, that it is not "improperly equivalent" to itself. Let me post a page from Dickson... $\endgroup$ – Will Jagy Jul 27 '16 at 18:38

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