Dirichlet's composition of a binary quadratic form multiple times: $f \circ f \circ f$.

Question

For an integer binary quadratic form $f(x,y) = A_1 x^2 + B_1 xy + C_1 y^2$ (or just $f = (A_1,B_1,C_1)$ for short), and another binary quadratic form $g = (A_2,B_2,C_2)$ with the same discriminant $(B_1^2 - 4A_1 C_1 = B_2^2 - 4A_2 C_2)$, Dirichlet showed that there exists integers $a,A,b,h$ such that there are equivalent forms $$f \sim F = (a,b,Ah),$$ $$g \sim G = (A,b,ah),$$ and their "composition" (equivalent to Guass's composition) is the class of forms given by $$(a,b,Ah) \circ (A,b,ah) \sim (Aa,b,h).$$

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My question is regarding calculating the composition of a form multiple times with itself. For example: $f \circ f \circ f$.

Will there always exist integers $a,b,h$ such that $$f \sim F = (a,b,a^2 h),$$ $$F^2 \sim F \circ F \sim (a,b,a^2 h) \circ (a,b,a^2 h) \sim (a^2,b,ah),$$ $$F^3 \sim F^2 \circ F \sim (a^2,b,ah) \circ (a,b,a^2 h) \sim (a^3,b,h) \ ?$$

I think this is equivalent to asking:

• Does every binary quadratic form have an equivalent quadratic form in the form $(a,b,a^2 h)$? (From which I think everything would follow.)

And for the more general case of composition more times: $f \circ f \circ ... \circ f$

• Does every binary quadratic form have an equivalent quadratic form in the form $(a,b,a^n h)$ for any positive integer $n$?

Answer 1

yes. In primitive $\langle a,b,c \rangle$ wishing to move to some $\langle a,b_1,c_1 a,$ we do need $\gcd(a,b) = 1$ Then the standard linear change $$ \left( \begin{array}{cc} 1& r \\ 0&1 \\ \end{array} \right) $$ gives us $\langle a,b + 2ar,c + br + a r^2 \rangle$ Our first step is to solve $$ c + br + a r^2 \equiv 0 \pmod a$$ or $$ c + br \equiv 0 \pmod a$$

Since there are constants with $ma + bn = 1,$ we see $bn \equiv 1 \pmod a$ Then $-cnb \equiv - c \pmod a,$ and we use $r = -cn$ Meanwhile, $b + 2ar \equiv b \pmod a $ and is still coprime with $a$

So far, we have $\langle a,b_1,a c_1 \rangle.$ Next step $\langle a, b_1 + 2at,ac_1 + b_1t + a t^2 \rangle.$ To keep the third coefficient divisible by $a$ we need to take $t = a u,$ so the result will be $\langle a, b_1 + 2a^2u,ac_1 + ab_1u + a^3 u^2 \rangle.$ We want $$ac_1 + ab_1u + a^3 u^2 \equiv 0 \pmod {a^2}, $$ $$ac_1 + ab_1u \equiv 0 \pmod {a^2}, $$ $$a(c_1 + b_1u) \equiv 0 \pmod {a^2}, $$ $$c_1 + b_1u \equiv 0 \pmod a. $$ We can solve this because $b_1$ is still invertible $\pmod a$

For example, with disriminant $-47,$ we see that the principal genus is cyclic by putting a $9$ in the middle,

$$\langle 1,9,32 \rangle.$$ $$\langle 2,9,16 \rangle.$$ $$\langle 4,9,8 \rangle.$$ $$\langle 8,9,4 \rangle.$$ $$\langle 16,9,2 \rangle.$$ NOTE $$\langle 32,9,1 \rangle\sim \langle 1,-9,32 \rangle\sim \langle 1,9,32 \rangle$$

It is harder to see things with indefinite forms. For discriminant $257$ we have class number exactly three, $$\langle 1,15,-8 \rangle.$$ $$\langle 2,15,-4 \rangle.$$ $$\langle 4,15,-2 \rangle.$$ For one thing, there are integer solutions to $x^2 + 15 xy - 8 y^2 = -1,$ which can be confirmed by finding the full Gauss-Lagrange cycle of the individual form,

jagy@gost:~/Desktop/Cplusplus$ ./indefCycle 1 15 -8

  0  form              1          15          -8


           1           0
           0           1

To Return  
           1           0
           0           1

0  form   1 15 -8   delta  -1     ambiguous  
1  form   -8 1 8   delta  1
2  form   8 15 -1   delta  -15
3  form   -1 15 8   delta  1     ambiguous            -1 composed with form zero  
4  form   8 1 -8   delta  -1
5  form   -8 15 1   delta  15
6  form   1 15 -8


  form   1 x^2  + 15 x y  -8 y^2 

minimum was   1rep   x = 1   y = 0 disc 257 dSqrt 16  M_Ratio  256
Automorph, written on right of Gram matrix:  
-33  -512
-64  -993
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