Algebraic relations between alternative representations of an integer in the form $p^2+3q^2$

Question

In the course of pondering this question I noticed that $4[(c+1)^3 – c^3]$ can be represented in the form $p^2 + 3q^2$ in three different ways:

$$4[(c+1)^3 – c^3] = 12c^2 + 12c + 4$$

$$= 1^2 + 3(2c+1)^2$$

$$= (3c+1)^2 + 3(c+1)^2$$

$$= (3c+2)^2 + 3c^2$$

This property also holds for some other integers not of the above form, for example:

$$52 = 7^2 + 3(1^2)$$

$$= 5^2 + 3(3^2)$$

$$= 2^2 + 3(4^2)$$

Other small integers with this property are $124$ and $172$. In each case the three representations are related by the identity:

$$p^2 + 3q^2$$

$$= \bigg(\dfrac{p+3q}{2}\bigg)^2 + 3\bigg(\dfrac{p-q}{2}\bigg)^2$$

$$= \bigg(\dfrac{p-3q}{2}\bigg)^2 + 3\bigg(\dfrac{p+q}{2}\bigg)^2\qquad(1)$$

This yields integers whenever $p$ and $q$ are both odd or both even. If any one representation is known, the other two can be readily calculated.

There are also integers which can be represented in the form $p^2+3q^2$ in six different ways, including $364$, $532$ and $868$. In each case the six ways can be divided into two groups of three, each exemplifying identity (1). For example:

$$364 = 19^2 + 3(1^2)$$

$$= 11^2 + 3(9^2)$$

$$= 8^2 + 3(10^2)$$

$$364 = 17^2 + 3(5^2)$$

$$= 16^2 + 3(6^2)$$

$$= 1^2 + 3(11^2)$$

Question: For integers representable in the form $p^2+3q^2$ in two groups of three ways, each group an instance of identity (1), is there any algebraic relation between the two groups (which would enable all the other five to be calculated if any one were known)?

Answer 1

(A partial answer. A complete answer would be relevant to a recent post of mine.)

I. Sum of cubes

Bailey mentioned he noticed $4\big((c+1)^3 – c^3\big)$ can be represented in the form $N = p^2 + 3q^2$ in three ways. That came from considering the equation,

$$a^3+b^3+c^3 = (c+m)^3$$

for $m=1$. However, if we allow rational $m$, then all the examples he gave can be used in such an equation. To illustrate,

Let $p^2+3q^2 = 364,$ and $r=\dfrac{34p+30q+17}{30}+11$ then,

$$\left(p+q+48\right)^3+\left(p-q+\tfrac{11}2\right)^3+r^3 = \left(r+\tfrac{75}2\right)^3$$

Let $p^2+3q^2 = 532,$ and $r=\dfrac{4p+3q+2}{3}+22$ then,

$$\left(p+q+42\right)^3+\left(p-q+10\right)^3+r^3 = \left(r+24\right)^3$$

Let $p^2+3q^2 = 868,$ and $r=\dfrac{11p+3q+2}{3}+106$ then,

$$\left(p+q+35\right)^3+\left(p-q+24\right)^3+r^3 = \left(r+3\right)^3$$

and similarly for $N = 52, 124, 172.$ But not all $N$ will do. The general procedure is discussed in this second post and involves an elliptic curve.

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II. Ellipses

Given an integer $N = p^2+3q^2$ expressible in three ways,

$$N = x_1^2+3y_1^2 = x_2^2+3y_2^2 = x_3^2+3y_3^2$$

it may be useful to let the $(x_k,y_k)$ be signed. So the given identity becomes,

$$N = (-p)^2+3(-q)^2 = \left(\frac{p+3q}2\right)^2+3\left(\frac{-p+q}2\right)^2 = \left(\frac{p-3q}2\right)^2+3\left(\frac{p+q}2\right)^2$$

such that $x_1 + x_2 + x_3 = y_1 + y_2 + y_3 = 0.$ These three $(x_k,y_k)$ will be the vertices of a triangle.

Assume the ellipse $p^2+3q^2-364=0.$ Starting with initial solution $(p,q) = (19,1)$ and using the identity above, we find three vertices. And the other initial solution $(p,q) = (17,5)$ will give three more vertices.

The blue dots give the first triangle, while the green dots give the second triangle. If instead we consider the six points as the vertices of a hexagon, then we have Pascal's Theorem,

$\hskip1.1in$

where the three pairs of the continuations of opposite sides meet on a straight line. So there is a relationship between the two triangles. But the question is if we can derive the second from the first.

In the first post I cited, I was able to derive from one triangle (blue vertices) another triangle (green vertices), resulting in a pair of distinct polynomial sextuplets that solve $x^3+y^3+z^3 = (z+1)^3$. For example, $594^3 + 1003^3 + 20154^3 = 20155^3.$

However, as pointed out by Eric Towers, one way to see the derivation was an appropriate rotation of the first triangle since connecting all six antipodes intersect at the center. But whether there is an appropriate rotation/transformation for Bailey’s two triangles remains to be seen.