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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.

$$x^2+y^2=u^2$$

$$y^2+z^2=v^2$$

$$x^2+z^2=w^2$$

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$x,y,z,u,v,w$ 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.

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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}

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