# Non-trivial automorph of an indefinite form

EDIT of the question by Will Jagy:

Is the binary quadratic form $x^2 - 3 y^2$ reduced? What are its proper integral automorphs?

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Unless of course we allow unimodular substitutions to have $-1$ determinant... –  Frustrated Jan 13 '12 at 0:08
But that wouldn't be in $\mathrm{SL}_2(\mathbb Z)$ –  Frustrated Jan 13 '12 at 0:18
Why do you delete the question? The answer here will not be of help to other users using the forum. Please do not delete questions for reasons that are now seen in the question section of this page! –  user21436 Jan 13 '12 at 2:56
@KannappanSampath the question was already deleted when I noticed it. I was curious enough to read the past edits, finding just the sort of thing that often confuses people. So I typed in a full answer. I think Frustrated is being a bit too hard on him/her self. The easiest version of reduction is the one I give...Do you want to edit in some version of the original question that makes Frustrated seem more curious and less "idiotic?" –  Will Jagy Jan 13 '12 at 3:03
@KannappanSampath I am Facebook friends with a first-year student at your school, named Koundinya Vajjha. He's funny. –  Will Jagy Jan 13 '12 at 3:06

As to reduced forms, see How to detect when continued fractions period terminates where, in particular, the reduced forms in the equivalence class of $x^2 - 3 y^2$ are $\langle 1,2,-2\rangle$ and $\langle -2,2,1\rangle.$

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$./Pell Input n for Pell 3 0 form 1 2 -2 delta -1 1 form -2 2 1 delta 2 2 form 1 2 -2 disc 12 Automorph, written on right of Gram matrix: 1 2 1 3 Pell automorph 2 3 1 2 Pell unit 2^2 - 3 * 1^2 = 1 ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$


which is to say

$$\left( \begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array} \right) \; \cdot \; \left( \begin{array}{cc} 1 & 0 \\ 0 & -3 \end{array} \right) \; \cdot \; \left( \begin{array}{cc} 2 & 3 \\ 1 & 2 \end{array} \right) \; = \; \left( \begin{array}{cc} 1 & 0 \\ 0 & -3 \end{array} \right)$$ and the process may be repeated, so that all $A^n$ with integer $n,$ either positive or negative, are automorphs. Here $$A \; = \; \left( \begin{array}{cc} 2 & 3 \\ 1 & 2 \end{array} \right)$$

In general, whether reduced or not, all the automorphs of a binary quadratic form $\langle a,b,c\rangle$ or $a x^2 + b x y + c y^2,$ with discriminant $\Delta = b^2 - 4 a c,$ are given by taking every solution of $$u^2 - \Delta v^2 = 4$$ and making the matrix $$A_{u,v} \; = \; \left( \begin{array}{cc} \frac{u - b v}{2} & -cv \\ av & \frac{u + b v}{2} \end{array} \right)$$ which works on $$\left( \begin{array}{cc} a & \frac{ b }{2} \\ \frac{ b }{2} & c \end{array} \right)$$

When $\Delta$ is positive and not a square, the number of solutions in integers to $u^2 - \Delta v^2 = 4$ is infinite.

NOTE: the matrux I call an automorph is the one written on the right of the matrix of the form itsef. This is what Duncan Buell does, in Binary Quadratic Forms, although you need to flip among pages to figure that out. H. E. Rose uses the matrix on the left in A Course in Number Theory.

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