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Let $q(x,y,z) = x^2 - by^2 - cz^2$ where $b,c$ are co-prime positive integers. Suppose that the binary quadratic form $f(x,y) = x^2 - by^2$ is irreducible. I am interested in counting integral points on the conic $C: q = 0$, but not with respect to the usual box height.

We say two pairs of integers $(x_1, y_1)$ and $(x_2, y_2)$ are related by the unit group of $\mathbb{Q}(\sqrt{b})$ if there exists a unit $u + v \sqrt{b}$ such that

$$\displaystyle (x_1, y_1) = (ux_2 + b vy_2, vx_2 + uy_2).$$

Define

$$\displaystyle N_q(B) = \# \{(x,y,z) \in \mathbb{Z}^3: q(x,y,z) = 0, |z| \leq B, $$

$$ \text{ and } (x,y) \text{ pairwise unrelated by the unit group} \}$$

From the outset, it is not even clear that $N_q(B)$ is finite. However, it is a classical theorem of Dedekind that the pairs $(x,y)$ satisfying $|x^2 - by^2| \leq B$ and such that you eliminate the unit group action by choosing only one pair from each orbit under the unit group has only finitely many solutions for any $B \geq 1$. Therefore, $N_q(B)$ is indeed finite.

What upper bounds are known for $N_q(B)$? I believe the bound $O(B \log \epsilon_b)$ should hold with an absolute constant, here $\epsilon_b = u_0 + v_0 \sqrt{b}$ is the smallest positive solution to the unit equation $x^2 - by^2 = 1$.

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I don't see that this need be tied to Pell's equation; I am taking your $b,c$ squarefree for ease. I have also picked the product so that there are no imprimitive forms of this discriminant. In comparison, if i had picked $5x^2 + 11 y^2,$ I would have had $ 2x^2 + 2xy + 28y^2,$ $4 x^2 + 2xy + 14 y^2,$ $8 x^2 + 6xy+8y^2$ to worry about.

With all this, we take $bx^2 + c y^2$ to be in the principal genus, here $17 x^2 + 89 y^2.$ Then the squares it represents (primitively) are squares of numbers that are primitively represented by its square roots in the class group of forms. The squares that are ruled out are those that share factors with the discriminant, $-4bc.$ For example, below we see plenty of even numbers or multiples of $17$ that are primitively represented by either $ 19 x^2 + 16 xy + 83 y^2$ or $ 38 x^2 + 22 xy + 43 y^2,$ but squares of numbers divisible by any of $2,17,89$ cannot be primitively represented by $17 x^2 + 89 y^2.$

I had a link to your inequality question before i went to the grocery store, then the car developed a coolant leak and I had to drop it at the mechanic and walk home. I think the point was that if $x^2$ is represented, then $x$ itself is represented by a different form of the same discriminant. Furthermore your $y,z$ come from Gauss duplication on that other form.

Not my day.

I will need to think some more about your forms outside the principal genus.

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jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./classGroup
Absolute value of discriminant? 
6052
Discr  -6052 = 2^2 * 17 * 89  class  number  16

 all  
    6052:  < 1, 0, 1513>    Square      6052:  < 1, 0, 1513>
    6052:  < 2, 2, 757>    Square      6052:  < 1, 0, 1513>
    6052:  < 11, -8, 139>    Square      6052:  < 34, 34, 53>
    6052:  < 11, 8, 139>    Square      6052:  < 34, 34, 53>
    6052:  < 17, 0, 89>    Square      6052:  < 1, 0, 1513>
    6052:  < 19, -16, 83>    Square      6052:  < 17, 0, 89>
    6052:  < 19, 16, 83>    Square      6052:  < 17, 0, 89>
    6052:  < 22, -14, 71>    Square      6052:  < 34, 34, 53>
    6052:  < 22, 14, 71>    Square      6052:  < 34, 34, 53>
    6052:  < 29, -26, 58>    Square      6052:  < 2, 2, 757>
    6052:  < 29, 26, 58>    Square      6052:  < 2, 2, 757>
    6052:  < 34, 34, 53>    Square      6052:  < 1, 0, 1513>
    6052:  < 37, -4, 41>    Square      6052:  < 2, 2, 757>
    6052:  < 37, 4, 41>    Square      6052:  < 2, 2, 757>
    6052:  < 38, -22, 43>    Square      6052:  < 17, 0, 89>
    6052:  < 38, 22, 43>    Square      6052:  < 17, 0, 89>

 squares  
    6052:  < 1, 0, 1513>
    6052:  < 2, 2, 757>
    6052:  < 17, 0, 89>
    6052:  < 34, 34, 53>

 fourths  
    6052:  < 1, 0, 1513>


Discriminant      -6052     h :   16     Squares :    4     Fourths :    1
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./squareprimitivego


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jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primgo
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2 
19 16 83
Discriminant  6052
Maximum number represented? 
1000
     19 =  19
     83 =  83
     86 = 2 *  43
    118 = 2 *  59
    127 =  127
    191 =  191
    206 = 2 *  103
    302 = 2 *  151
    319 = 11 *  29
    323 = 17 *  19
    383 =  383
    407 = 11 *  37
    451 = 11 *  41
    478 = 2 *  239
    599 =  599
    638 = 2 * 11 *  29
    647 =  647
    671 = 11 *  61
    718 = 2 *  359
    727 =  727
    814 = 2 * 11 *  37
    859 =  859
    863 =  863
    902 = 2 * 11 *  41
    919 =  919
    967 =  967
    982 = 2 *  491
--------------------------------------------------------------------------
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primgo
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2 
 38 22 43
Discriminant  6052
Maximum number represented? 
1000   
     38 = 2 *  19
     43 =  43
     59 =  59
    103 =  103
    151 =  151
    166 = 2 *  83
    239 =  239
    254 = 2 *  127
    319 = 11 *  29
    359 =  359
    382 = 2 *  191
    407 = 11 *  37
    451 = 11 *  41
    491 =  491
    563 =  563
    638 = 2 * 11 *  29
    646 = 2 * 17 *  19
    671 = 11 *  61
    731 = 17 *  43
    739 =  739
    766 = 2 *  383
    814 = 2 * 11 *  37
    883 =  883
    902 = 2 * 11 *  41
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ 
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jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./squareprimitivego
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2 
17 0 89
Discriminant  -6052

Maximum number represented? 
1000000

         361 = 19^2
        1849 = 43^2
        3481 = 59^2
        6889 = 83^2
       10609 = 103^2
       16129 = 127^2
       22801 = 151^2
       36481 = 191^2
       57121 = 239^2
      101761 = 11^2 * 29^2
      128881 = 359^2
      146689 = 383^2
      165649 = 11^2 * 37^2
      203401 = 11^2 * 41^2
      241081 = 491^2
      316969 = 563^2
      358801 = 599^2
      418609 = 647^2
      450241 = 11^2 * 61^2
      528529 = 727^2
      546121 = 739^2
      737881 = 859^2
      744769 = 863^2
      779689 = 883^2
      844561 = 919^2
      935089 = 967^2

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  • $\begingroup$ Sorry to hear about your day! In any case, the issue here is that I need to sort with respect to one variable only, and so this creates challenges when attempting asymptotic estimates. $\endgroup$ – Stanley Yao Xiao Aug 19 '16 at 0:53

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