Finding a rational (or integral) parametrization for a system of (homogeneous?) quadratic forms?

Question

The solution of the $3 \times 3$ magic square of squares involves finding $8$ arithmetic progressions of squares (APS) using $9$ integers. I'm looking at a much smaller portion of the problem: $3$ APSs on $6$ integers.

The general APS $(a^2, b^2, c^2)$ has this parametrization:

\begin{cases} a = |2pq - p^2 + q^2| \\ b = p^2 + q^2 \\ c= 2pq + p^2 - q^2 \end{cases}

For any $p,q \in \mathbb{N}, p > q$. Primitive APSs have $(p,q) = 1$.

The system I'm examining looks like this:

\begin{align} B, I, D &= |2ab - a^2 + b^2| \ ,\ a^2 + b^2 \ ,\ 2ab + a^2 - b^2 \\ B, G, F &= |2cd - c^2 + d^2| \ ,\ c^2 + d^2 \ ,\ 2cd + c^2 - d^2 \\ D, E, F &= n(|2fg - f^2 + g^2|) \ ,\ n(f^2 + g^2) \ ,\ n(2fg + f^2 - g^2) \\ \end{align}

with $a>b, c>d, f>g$, and where $(B,D,E,F,G,I)$ represent the positions in a standard lettered $3 \times 3$ square. EDIT: Earlier I had thought first two APSs had to be primitive, but the third need not be. I've now found examples where none are primitive, but they're very rare.

The system gives the following system of quadratic equalities from $B,D,F$:

\begin{align} |2ab - a^2 + b^2| &= |2cd - c^2 + d^2| \\ 2ab + a^2 - b^2 &= n(|2fg - f^2 + g^2|) \\ 2cd + c^2 - d^2 &= n(2fg + f^2 - g^2) \\ \end{align}

There appear to be an infinite (though sparse) number of solutions to these equalities, and I suspect that we can parametrize the system. Given we have three equalities in seven variables, I assume we'll need four integer parameters $(p,q,r,s)$ to determine the solutions.

I've used information from this question and an answer to a different question in an attempt to parametrize just the third equation, but my attempts have been blotchy and don't directly give any of the values; for instance, from one known set of values (see footnote), I was able to achieve:

$$\frac{d}{c+d} = \frac{kp^2 - 2mpq - 2kq^2}{p^2 + 2q^2}, \frac{f+g}{c+d} = \frac{2mq^2 - mp^2 - 4kpq}{p^2 + 2q^2}$$

where $(k, m) = (\frac{4}{19}, \frac{23}{19})$. Unfortunately, this gives the original solution using $(p,q) = (-152, 437)$, returning only the rational values, and I'd like to actually get $c, d, f, g$ rather than fractions involving them.

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I don't expect anyone is likely to be able to provide a full answer, and possibly not even a partial one; I'm hoping some of you might be able to provide pointers, links, ideas, etc. You can assume I've read the vast majority of information and papers at the best-known site about the problem. I haven't worked with quadratic forms or fields, though I've picked up a few bits here and there. The absolute values seem like a major obstacle, which is why I started with the third equation alone. While finishing this up I've come across this question, which seems to be possibly related; some of the same values there are ones I'm using, but I haven't had a chance to fully digest it yet.

Thanks for any thoughts!

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Footnote: the solution I'm using as a basis is this one: there are plenty more where that came from though.

\begin{align} (B,I,D,G,F,E) &= (89,149,241,191,269,329) \\ (a,b,c,d,f,g,n) &= (10, 7, 15, 4, 13, 10, 1) \end{align}

A better-known solution has $(B,I,D,G,F,E) = (23,205,289,373,425,527)$, which is notable because it is the only example where a seventh integer, $A=565$, generates the largest-known (i.e., $7/9$) partial order-$3$ magic square of squares.

Answer 1

The system,

\begin{align} |- a^2 + 2ab + b^2 |&=| - c^2 + 2cd + d^2 |\\ + a^2 + 2ab - b^2 \;&=\; - f^2 + 2fg + g^2 \\ + c^2 + 2cd - d^2 \;&=\; + f^2 + 2fg - g^2 \end{align}

can be solved by an elliptic curve. Define,

\begin{align} c &= p(m+1)+m q\\ d &= -p+(m-2)q\\ f &= p(m-1)+m q\\ g &= p+(m+2)q \end{align}

then it is solved by,

\begin{align} a &= \frac{2(p^2+2pq+2q^2)\sqrt{b^2-2(p^2-2q^2)}}{\sqrt{b^2(p^2-2q^2)+4(p^2+2pq+2q^2)^2}}\\ m &= \frac{b\sqrt{b^2-2(p^2-2q^2)}}{\sqrt{b^2(p^2-2q^2)+4(p^2+2pq+2q^2)^2}} \end{align}

for three free parameters $(b,p,q)$. But since rational solutions are desired, then $(b,p,q)$ must satisfy,

$$\Big(b^2-2(p^2-2q^2)\Big)\Big(b^2(p^2-2q^2)+4(p^2+2pq+2q^2)^2\Big) = z^2$$

with initial (but trivial) rational points $b = 2p$ and $b = \frac{p^2+2pq+2q^2}q$. However, this quartic in $b$ is birationally equivalent to an elliptic curve and, from these trivial polynomial solutions, one can derive infinitely many non-trivial ones.

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Example.

This is the special case $a=g$. Define,

\begin{align} a &= p^2 - 4pq + 6q^2 + z\\ b &=\frac{(-p + 3q)(3p^2 - 4pq + 2q^2 + z)}q\\ c &=\frac{-3p^3 + 15p^2q - 26pq^2 + 18q^3 - (p - q)z}q\\ d &=-5p^2 + 16pq - 6q^2 - z\\[3pt] f &=\frac{+3p^3 - 11p^2q + 6pq^2 + 6q^3 + (p - q)z}q\\ g &= a = p^2 - 4pq + 6q^2 + z \end{align}

where $(p,q)$ must satisfy,

$$9p^4 - 48p^3q + 92p^2q^2 - 96pq^3 + 36q^4 = z^2$$

The point $(p,q) = (2,3)$ yields, after removing common factors, the first solution of the OP,

$$(a,b,c,d,f,g) = (10, 7, 15, 4, 13, 10)$$

while the point $(p,q) = (14,3)$ yields,

$$(a,b,c,d,f,g) = (91, -300, -442, -161, 482, 91)$$

and infinitely more.