Finding or parametrizing integer solutions to $pq(p^2-q^2)=rs(r^2-s^2)$

Question

Background: The order-3 magic square of squares problem (MSS3) is a well-known open problem that involves finding eight separate arithmetic progressions of three squares (APSs). In particular, two sets of three APSs must have equal congrua--the difference between successive terms.

There's a well-known parametrization for APSs and their congrua. For a progression $i^2, j^2, k^2$ with congruum $d$, we have:

$$ \begin{cases} i = \lvert 2mn - m^2 + n^2 \rvert \\ j = m^2 + n^2 \\ k = 2mn + m^2 - n^2 \\ d = 4mn(m^2 - n^2) \end{cases} $$

Here $m, n \in \mathbb{Z}, m > n$. If we desire primitive APSs (and we usually do), we further restrict our choices such that $m$ and $n$ have opposite parity, and $(m,n)=1$. From the equalities above, finding two primitive APSs with the same congruum requires finding a solution to the equation in the title,

$$pq(p^2 - q^2) = rs(r^2 - s^2) \tag{1}$$

(There is also the possibility that a scaling factor of $t^2$ inserted on either side will create another solution, but for the moment I'm setting that off to the side; the same method likely can't account for both.)

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Stuff I've Tried or Seen

(A) Brute Force: Letting Sagemath run a bunch of calculations is a thing we can do, look at the output and hope to find patterns. It turns out that this is hard; if there's any pattern, it's not something that pops out. However, what's interesting is that for $m < 2000$, there are only a few hundred matching pairs (out of close to a million primitive APSs). That's really surprising to me; I thought they'd be less rare. The rarity likely makes trying to find the patterns more difficult.

Surprisingly, there is exactly one set of three APSs with the same congruum within the same data, suggesting that triples of that sort are rare indeed. (Note: I'm aware that the values with $m<2000$ are too low to be part of the solution of the MSS3 problem; I'm trying to find patterns to use in arguing against the existence of a solution.)

(B) A Geometric Interpretation: There are simple geometrical interpretations of both the Pythagorean triples (PTs) and APSs:

For a PT, we know $a^2 + b^2 = c^2$, and we can divide through by $c$, letting $x=a/c, y=b/c$, and draw a unit circle. A line through $(-1,0)$ with a rational slope intersects the unit circle at the rational point $(a/c, b/c)$.

For the APS above, $i^2 + k^2 = 2j^2$ (as the differences between the squares are equal); we divide through by $j$ and draw the circle $x^2 + y^2 = 2$. A line through $(-1, -1)$ with a rational slope intersects this circle at the rational point $(i/j, k/j)$.

If the same rational slope is chosen for both, you have an APS and a PT that correspond to one another: $j=c, i=a-b, k=a+b$. Here at Desmos is what I think is a good visualization of these interpretations.

How to apply this to equation $(1)$? Well, we can divide through by $s$, letting $x = p/s, y = q/s, z = r/s$, giving us:

$$xy(x^2 - y^2) = z(z^2 + 1) \tag{2}$$

But because this equation is a quartic, it's not enough to have just one rational point to start from. We need at least two, so far as I can tell. Not to mention we're dealing with a surface now and not a curve. So it fails to help us the way we would like. I can find some solutions using Sage, but not as fast as brute force finds them.

(C) Prior Partial Answers: This equation has been discussed at least once before on MSE, here. The only answer there converts $(1)$ above into the equivalent:

$$(p + qi)^4 + (r - si)^4 = (p - qi)^4 + (r + si)^4 \tag{3}$$

The answerer says this can be solved with elliptic curves, says there are an infinite number of parametrizations, gives a single one... and does not explain how to deal with any of that. Expanding and simplfying simply gives equation $(1)$ back.

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So... has anyone else worked with this equation? And/or can someone explain how $(3)$ can be manipulated into an elliptic curve?

Answer 1

Since I'm the guilty "answerer" alluded to in the link from 7 years ago, we'll try to provide more details this time.

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I. Generating infinitely many parametrizations

Eq.3 of the OP can be generalized as,

$$(p + \sqrt{h}q)^4 + (r - \sqrt{h}s)^4 = (p - \sqrt{h} q)^4 + (r + \sqrt{h}s)^4$$

Or equivalently,

$$p q(p^2 + h q^2) = r s(r^2 + h s^2)$$

But we have the general identity,

$$a b(a^2 + h b^2) - c d(c^2 + h d^2) = \big(p q(p^2 + h q^2) - r s(r^2 + h s^2)\big)\big(m^3q^2 - n^3s^2\big)^4$$

where $(a,b,c,d)$ are,

\begin{align} a &= +m^3p q^2 - 3m^2n q r s + 2n^3p s^2\\ b &= +q(m^3q^2 - n^3s^2)\\ c &= -2m^3q^2r + 3m n^2p q s - n^3 r s^2\\ d &= +s(m^3q^2 - n^3s^2)\\[4pt] m &= 3p^2 + h q^2\\ n &= 3r^2 + h s^2 \end{align}

which is true for any $h$ and $(p,q,r,s)$. So if there are initial rational points $(p,q,r,s)$ such that $RHS=0$, this yields new points $(a,b,c,d)$ such that $LHS = 0.$ For example, an initial parametrization of deg-$(4,5)$ for $h=-1$ is,

$$(p,\;q,\;r,\;s) = (v^5-2v,\;v^5+v,\;-2v^4+1,\;v^4+1)$$

Substituting these into the formulas and removing common factors yields deg-$(16,17)$ polynomials,

\begin{align} a &= 2v(1 + v^4)(-1 + 2v^4)(-16 + 13v^4 + 2v^8)\\ b &= v(8 - 184v^4 + 345v^8 - 184v^{12} + 8v^{16})\\ c &= 2(-2 + v^4)(1 + v^4)(-2 - 13v^4 + 16v^8)\\ d &= (8 - 184v^4 + 345v^8 - 184v^{12} + 8v^{16}) \end{align}

Letting this 2nd parameterization be the new $(p,q,r,s)$ will yield a third parameterization of deg-$(64,65)$, and so on. Thus, the general identity is equivalent to,

$$x_1^4+x_2^4-x_3^4-x_4^4=(y_1^4+y_2^4-y_3^4-y_4^4)\,P(y_k)^4$$

which is the $4$th power version,

$$x_1^3+x_2^3-x_3^3-x_4^3=(y_1^3+y_2^3-y_3^3-y_4^3)\,P(y_k)^3$$

found in this post.

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II. Elliptic curve

Alternatively, we can use the same technique that Euler used for $x_1^4+x_2^4 = y_1^4+y_2^4$. Given,

$$pq(p^2-q^2) = rs(r^2-s^2)$$

Let $(p,\;q,\;r,\;s) = (x, \; ay,\; bx,\; y)$ and it becomes,

$$(a - b^3) = (a^3 - b)y^2/x^2$$

Multiply both sides by $(a^3 - b)$, and the $RHS$ becomes a square,

$$(a^3 - b)(a - b^3) = z^2$$

An initial solution is,

$$a = v,\quad b =\frac{1-2v^4}{-2v+v^5}$$

A quartic polynomial to be made a square with a non-trivial rational point is birationally equivalent to an elliptic curve. From this initial point, an infinite more can then be found.