Is there a complete answer to this problem? I have found Saunderson's answer, but I believe it is missing a few answers. The problem states:

$a^2+b^2=d^2 \\ a^2+c^2=e^2 \\ b^2+c^2=f^2$

Saunderson proves the answer is

$a=y(4x^2-z^2)\\ b=x(4y^2-z^2)\\ c=4xyz$

where $x,y,z$ is the Pythagorean triple $x^2+y^2=z^2$. But this skips answers like $(85,132,720)$,$(132,351,720)$, etc.

The complete Saunderson proof is here: https://play.google.com/books/reader?id=1NI_AQAAMAAJ&printsec=frontcover&output=reader&hl=en&pg=GBS.PA429

The solutions are also known as Euler Bricks.

Also (since I don't think a complete answer exists), do any of you have suggestions on how to find one?

  • $\begingroup$ Is it easy to see that your examples aren't covered by Saunderson's formulas? $\endgroup$ – Gerry Myerson May 9 '15 at 12:39
  • $\begingroup$ Depends what you mean by easy to see. I run programs, and that is how I know. $\endgroup$ – Seth Kitchen May 11 '15 at 20:48

The four smallest Euler bricks are,

$$44, 117, 240$$

$$85, \color{brown}{132}, \color{brown}{720}$$

$$88, 234, 480$$

$$\color{brown}{132}, 351, \color{brown}{720}$$

Notice that the second and fourth bricks share two terms. There are an infinite number of such Euler brick pairs. Let $u^2+v^2 = 5w^2$, then,

$$a,b = (u^2-w^2)(v^2-w^2),\; 4uvw^2$$


$$c = 2uw(v^2-w^2)\;\;\color{brown}{or}\;\; 2vw(u^2-w^2)$$

This is a second parameterization in terms of quadratic forms. Bremner in "The Rational Cuboid and a Quartic Surface" showed there are many, many identities. As Bremner points out in the last page, "...there will be a rational parametrization of every even degree greater than or equal to six...".

It is then doubtful there is a single polynomial identity that covers all solutions.

  • $\begingroup$ Do you have any suggestions about how to go about covering all solutions? $\endgroup$ – Seth Kitchen May 11 '15 at 18:55

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