# Counting integral points on a diagonal conic

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 $x^2 - by^2 = 1$.

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$------------------------------------------------------------------ jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./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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• 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. – Stanley Yao Xiao Aug 19 '16 at 0:53