Cuboid nearest to a cube.
While answering this question, euler bricks: way to calculate them? I noticed one result was not too far from cube shaped, and wondered if there was a more cubic cuboid.
$$x^2+y^2=u^2$$ $$y^2+z^2=v^2$$ $$x^2+z^2=w^2$$
$x,y,z,u,v,w$ positive integers, and $x<y<z$
The result I noticed was $(240,252,275)$, and decided to use $\alpha=\large \frac{z^2}{xy}$ as a measure of nearness to a cube. For $(240,252,275)$ we have $\alpha=1.25041336$
Diagram: https://en.wikipedia.org/wiki/Euler_brick#/media/File:Euler_brick_examples.svg
Despite a fair bit of calculation, I can only find one more cubic cuboid: $$(1008,1100,1155)$$
This has $\alpha=1.203125$ and is produced from the following solution generator using $(240,252,275)$, “If $(x,y,z)$ is a solution, then $(xy,xz,yz)$ is also a solution”.
My questions. Is there a better measure of nearness to a cube than $\alpha= \large\frac{z^2}{xy}$ ?
Is there a better solution than $(1008,1100,1155)$ ?
Thank you.
Answer to the $2$nd question: yes, there are cubic cuboids with rather less measure.
If use measure $\alpha = \dfrac{z^2}{xy}$, then the best one currently known for me (see the table below) has $\alpha \approx \color{red}{1.0352}$.
Here are few noteworthy examples:
\begin{array}{|r|c|} \hline (x,y,z) & \alpha \\ \hline (2\:278\:100,\; 2\:423\:952,\; 2\:564\:661) & \approx 1.191 \\ (4\:160\:772,\; 4\:540\:525,\; 4\:717\:440) & \approx 1.178 \\ (14\:358\:336,\; 15\:041\:873,\; 15\:526\:440) & \approx 1.116 \\ (43\:875\:188,\; 44\:127\:291,\; 46\:181\:520) & \approx 1.102 \\ (5\:122\:780,\; 5\:245\:200,\; 5\:288\:547) & \approx 1.0409 \\ (15\:301\:440,\; 15\:748\:920,\; 15\:798\:809) & \approx 1.0358 \\ (108\:192\:528,\; 109\:141\:700,\; 110\:562\:771) & \approx 1.0352 \\ \hline \end{array}