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I have been looking into the problem of the magic square of squares, or more specifically, the magic hourglass of squares, like so:

$a^2$ $b^2$ $c^2$

$ $ $ $ $ $ $ $ $ $ $d^2$

$e^2$ $f^2$ $g^2$

Where both rows, the column, and the two diagonals add to the same total. It turns out, when you pick values of a, d, g which satisfy this, and try to find ways to make three of the four other terms work, you end up with an elliptic curve, which has rank 2. This can be parametrized, and taking one of the generators, you end up with:

$a=(m^2-2mn-n^2)(m^2+n^2)(5m^2-4mn+n^2)$,

$b=(m^2+2mn-n^2)(7m^4-16m^3n+6m^2n^2-n^4)$,

$c=m^6+24m^5n-13m^4n^2-16m^3n^3+19m^2n^4-8mn^5+n^6$

$d=(m^2+n^2)^2(5m^2-4mn+n^2)$

$f=m^6-26m^5n+17m^4n^2+12m^3n^3-17m^2n^4+6mn^5-n^6$

$g=(m^2+2mn-n^2)(m^2+n^2)(5m^2-4mn+n^2)$

In order for this to be a true hourglass of squares, the $e^2$ term must also work, yielding the equation:

$e^2=49m^{12}-128m^{11}n-298m^{10}n^2+320m^9n^3+1071m^8n^4-1856m^7n^5+1140m^6n^6-64m^5n^7-321m^4n^8+192m^3n^9-42m^2n^{10}+n^{12}$

Or, after switching variables and dehomogenizing:

$y^2=49x^{12}-128x^{11}-298x^{10}+320x^9+1071x^8-1856x^7+1140x^6-64x^5-321x^4+192x^3-42x^2+1$

Solving this equation is quite a bit above my paygrade as an amateur mathematician.

Are there any rational solutions to this equation, other than the trivial $x=0, 1, -1$, or the point at infinity? Any new solution would immediately yield a solution to the previously unsolved magic hourglass of squares!

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  • $\begingroup$ First thought: have you tried working modulo small numbers? Mod 3, 8 and 5 all come to mind as being worth a quick try to prove impossibility from. (Though this will admittedly only give integer solutions, not rational.) $\endgroup$
    – Steven Stadnicki
    Commented Nov 25, 2021 at 5:09
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    $\begingroup$ The existence of a few trivial integer solutions seems to rule out the modular approach, especially for small bases such as 3, 5, or 8. $\endgroup$
    – Thomas
    Commented Nov 25, 2021 at 5:34
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    $\begingroup$ Every m,n combination in mod 3, mod 5, and mod 8 works to make e^2 a square, so that is out. $\endgroup$
    – Thomas
    Commented Nov 25, 2021 at 6:38
  • $\begingroup$ It's not clear to me from the presentation, where what's already known stops, and what's your new work, if any, begins. If the $y^2=49x^{12}+\cdots$ equation is already in the literature, and no nontrivial solutions are known, that suggests that so far the problem is above everyone's pay grade, not just yours. $\endgroup$
    – Gerry Myerson
    Commented Nov 26, 2021 at 2:21
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    $\begingroup$ I'd be worried, Thomas, that you might be reinventing the wheel. As noted in other comments, Andrew Bremner and others have studied this problem before. You might get some ideas from their work, or you might find they've already been where you are. $\endgroup$
    – Gerry Myerson
    Commented Nov 27, 2021 at 23:00

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