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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.

$$x^2+y^2=u^2$$ $$y^2+z^2=v^2$$ $$x^2+z^2=w^2$$

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

Old Peter
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  • Just off hand, it seems that $z^2/(xy)$ is not the best since the value of $y$ effects it ``too much". I noticed that if $y=x+1$, then $\alpha$ is larger than if $y=z-1$ (in other words, it's better for $y$ to be close to $z$ than close to $x$). Perhaps $z/x$ is better, or perhaps there is a reason the value of $y$ should effect $\alpha$ in this way? – Matthew Conroy May 15 '17 at 20:18
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    Forgive my ignorance. Are you talking about an Euler brick, a perfect cuboid or something else entirely? – Mr. Brooks May 15 '17 at 21:36
  • One measure is neither better nor worse than another, unless you introduce some sort of measure for measures, so as to measure which of them is better. As for the cuboids, I guess you can get arbitrarily close to a cube. – Ivan Neretin May 16 '17 at 07:29
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    @Mr.Brooks Yes, an Euler brick or cuboid. Not a perfect cuboid, as I don't know of any. – Old Peter May 16 '17 at 09:14
  • @MatthewConroy I'm open to ideas, for any method that's based on shape, but not size. – Old Peter May 17 '17 at 18:29
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    @OldPeter How does my idea for $z/x$ strike you? – Matthew Conroy May 17 '17 at 20:12
  • @MatthewConroy It strikes me as very interesting, and could well add insight into these solutions; I will have to investigate. However, as a lot of work has been done, I’m reluctant to change now. Thank you for your help. – Old Peter May 18 '17 at 10:09

1 Answers1

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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}

Oleg567
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    @OldPeter, answer is updated now with better examples. – Oleg567 May 17 '17 at 05:56
  • Thank you so very much for all these splendid results! I has noticed the results earlier, and the minor issue you corrected, but it’s been one of those “too-much-to-do” days. – Old Peter May 17 '17 at 18:46