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Saunderson found a parametric solution giving Euler bricks:

Let $(x,y,z)$ be a pythagoran triple, then we get an Euler brick $(a,b,c)$ with $a=x(4y^2-z^2)$, $b=y(4x^2-z^2)$, $c=4xyz$.

It is claimed, that the parametric solution is not giving all possible Euler bricks- but why? (https://mathworld.wolfram.com/EulerBrick.html)

Can you think of any example? And maybe a proof why that example is not given by the parametric solution?

Tito Piezas III
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xbut123
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2 Answers2

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With $x^2+y^2=z^2$, $a=x(4y^2-z^2)$, and $b=y(4x^2-z^2)$, it is easy to show that, for the face diagonal $d_{ab}$, one has ${d_{ab}}^2=a^2+b^2=z^6$, hence $d_{ab}=z^3$ is necessarily a perfect cube.

If such parametrization could give all Euler Bricks, then all Euler Bricks would have a perfect cube in one of its face diagonals. But it is not the case. Any Euler Brick without a perfect cube in its face diagonals is not given by such parametrization.

Marnes
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The short answer is because there are infinitely many parameterizations, not just one, with Saunderson's being the simplest. There are at least two methods to generate an infinite sequence of formulas: Euler's and Lenhart's.

I. Euler's Method

$$(a,b,c) = \left(\frac{p^2 - 1}{2p},\; \frac{q^2 - 1}{2q},\; 1\right)$$

This makes $a^2+c^2$ and $b^2+c^2$ as squares where denominators can be cleared later. It remains to make $a^2+b^2$ a square. One choses $(p,q)$ such that,

$$(p^2 - 1)^2q^2+p^2(q^2 - 1)^2 = t_1^2$$

Thus a quartic polynomial to be made a square. Given a rational point, this can be birationally treated as an elliptic curve, hence we can generate subsequent points from the initial one. Given Condition I or $x^2+y^2=z^2$ the Pythagorean triple $(x,y,z)$, then,

\begin{align} p_1&=\frac{2y}{z},\quad q_1=\frac{z}{2x}\\[5pt] p_2&=\frac{2y}{z},\quad q_2=\frac{ y(2x + z)}{(x + z)(2x - z)}\end{align}

and so on for infinitely many $(p_n,q_n)$.

Example. Let $(x,y,z)=(3,4,5)$, then

$$(a,b,c) = (117, 44, 240)\\ (a,b,c) = (429, 2340, 880)$$

and so on. Of course, the formula for $x^2+y^2=z^2$ is well-known, so we can have used that instead to get a polynomial parameterization. The variables $(p_1, q_1)$ in fact yield Saunderson's formula which is the simplest,

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

Again let $(x,y,z) = (3,4,5)$, then $(a,b,c) = (117, 44, 240)$, same as Euler's above.

II. Lenhart's Method

(See this MSE post.)

Tito Piezas III
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