Talk:Euler brick

old thread

I believe the conditions on the edges given for a perfect cuboid apply to all Euler bricks. Also a perfect cuboid must have edges divisible by 7 and 19. Ringman8567 (talk) 18:13, 28 January 2008 (UTC)

Doesn't the following site have a proof that perfect cuboids do not exist?: [1] —Preceding unsigned comment added by 12.2.219.70 (talk) 18:09, 2 March 2008 (UTC)

• Has the proof been published in a journal? I don't think it has. I would assume that if it were a valid proof, it would have been published.TheRingess (talk) 06:15, 17 July 2008 (UTC)

Facts?

Some interesting facts about a perfect cuboid.

• 2 edges must be even and 1 edge must be odd (for a primitive perfect cuboid).
• 1 edge must be divisible by 4 and 1 edge must be divisible by 16
• 1 edge must be divisible by 3 and 1 edge must be divisible by 9
• 1 edge must be divisible by 5
• 1 edge must be divisible by 11

I think this must be a mistake. See this page on f2.org. It says, those are "Some properties of primitive Euler bricks"! In fact, primitive Euler bricks are not same as Euler bricks or perfect cuboid! --Cnchina (talk) —Preceding comment was added at 01:38, 17 July 2008 (UTC)

Actually, if the perfect cuboid exists, it will be a primitive Euler Brick, so it does in fact share the same properties. If I remember it right, a primitive Euler brick is one with sides whose gcf is 1. So the facts are correct.TheRingess (talk) 06:13, 17 July 2008 (UTC)

It's not true that all perfect cuboids are primitive Euler Bricks. If (a,b,c) is a perfect cuboid, then (2a,2b,2c) is also a perfect cuboid, but isn't primitive. Perhaps you mean "If a perfect cuboid exists, then a primitive perfect cuboid also exists"? This would be true, since any non-primitive could be converted to a primitive by dividing out the gcd. Staecker (talk) 11:23, 28 July 2009 (UTC)

The above is incomplete.

• Because one leg of a Pythagorean triangle must be divisible by 4, then one of the triangles in an Euler block must have BOTH legs divisible by 4, so the hypotenuse is too. Dividing out those 4s, we still have that one leg of the reduced triangle must be divisible by 4, so the original must be divisible by 16. So the correct statement is that 3 edges (2 legs and 1 hypotenuse) must be divisible by 4, and one of those (a leg) must actually be divisible by 16. This means that the product abcdef must be divisible by 256 = 2^8, and the product abc by 64.
• Similarly, because one leg of a Pythagorean triangle must be divisible by 3, one whole triangle must be divisible by 3. So 3 edges (2 legs and 1 hypotenuse) must be divisible by 3, and one of those legs must actually be divisible by 9. The product abcdef must be divisible by 81 = 3^4, and the product abc by 27.
• Because one edge of a Pythagorean triangle must be divisible by 5 (but it can be either a leg (5 12 13) or the hypotenuse (3 4 5)), at least 2 edges total must be divisible by 5. This is easy to see since any one edge can only be shared by two triangles, so it takes a second edge to satisfy the third triangle. (And they must not share a triangle, so one must be a leg and the other the opposite hypotenuse.) So abcdef is divisible by 25 = 5^2.

Taking all these plus the factor of 11, we get that abcdef must be divisible by 5702400 = 2^8 * 3^4 * 5^2 * 11 and abc must be divisible by 1728 = 2^6 * 3^3. I'll go ahead and try to make the appropriate changes soon without going into the gory details, but there's a question of the reference matching, so I'll check that first. Howard Landman (talk) 19:46, 12 June 2017 (UTC)

a Euler brick vs. an Euler brick

I believe the name Euler is pronounced the same as "oiler", so the indefinite article should be "an", not "a". Comments?—GraemeMcRae^talk 18:42, 10 April 2009 (UTC)

This page discusses a Perfect parallelepiped in the text:

"A perfect cuboid is the special case of a perfect parallelepiped with all right angles. In 2009, a perfect parallelepiped was shown to exist,[4] answering an open question of Richard Guy. Solutions with only a single oblique angle have been found."

A Perfect parallelepiped has been found. You can get a layman's discussion at http://www.science20.com/alpha_meme/infinitely_improbable_coincidences

and the paper by it's discoverers Jorge F. Sawyer; Clifford A. Reiter at http://www.ams.org/journals/mcom/0000-000-00/S0025-5718-2010-02400-7/home.html

This is the first time I have posted on Wikipedia and I apologize in advance if this is out of order. —Preceding unsigned comment added by 24.23.250.181 (talk) 21:38, 27 August 2010 (UTC)

In other words...

$a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\ =\ c^{2}\ =\ {\frac {b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}{2}}$

Now, if only someone were to show that the sum of three non-zero perfect squares can never be both a perfect square itself, and the semi-sum of other three non-zero perfect squares at the same time... — 79.113.221.97 (talk) 01:17, 20 March 2013 (UTC)

If you are referring to the above discussion about primitive vs non-primitive Euler Bricks, your equation should be something like $a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\ =\ c^{2}\ =\ {\frac {b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}{4}}$ where $b_{1}=2*a_{1},b_{2}=2*a_{2},b_{3}=2*a_{3}$. Nothing to prove there.TheRingess (talk) 04:09, 20 March 2013 (UTC)

No, that's not what I'm talking about. When you add the squares of the three diagonals together, you get twice the sum of the squares of the three edges, which itself is the square of the main diagonal. — 79.113.221.97 (talk) 04:22, 20 March 2013 (UTC)

Okay. Maybe someone will prove it some day.TheRingess (talk) 07:14, 20 March 2013 (UTC)

No, this is not equivalent -- you've lost important information relating your as and bs. For example, try "edges" of lengths 1,2,2, "diagonals" of lengths 1,1,4, and "space diagonal" of length 3: these numbers satisfy your equations but don't yield an Euler birch. --JBL (talk) 14:07, 20 March 2013 (UTC)

I never said that they are "equivalent"... merely that the former implies the latter. So, if the latter would be proven untrue, so would that which gave birth to it. Reduction to the absurd. — 79.113.214.121 (talk) 16:41, 20 March 2013 (UTC)

sigh . Anonymous, Joel B just showed you that your statement as written is false. He showed you that the sum of 3 non zero perfect squares 1 + 4 + 4 = 9, and 9 is a perfect square, can also be written as the semi-sum of other three non zero perfect squares (9 = (1 + 1 + 16)/2). Therefore your theoem as written is false. There's no reduction to the absurd, nothing like that, merely a counter example that disproves the theorm.TheRingess (talk) 04:12, 21 March 2013 (UTC)

OK. Thanks. — 79.113.214.121 (talk) 05:56, 21 March 2013 (UTC)

Unclear sentence

Currently at the end of the article is the sentence

Solutions [for a perfect parallelepiped] with only a single oblique angle have been found.

Taken literally this cannot be true, since each face is a quadrilateral and no quadrilateral has only a single oblique angle. Can someone clarify in the article? Loraof (talk) 16:40, 30 March 2015 (UTC)

Proven?

https://arxiv.org/abs/1506.02215 — Preceding unsigned comment added by 188.146.74.175 (talk) 12:53, 12 March 2017 (UTC)

arXiv has no peer review and is not by itself an acceptable source for Wikipedia. See Wikipedia:Reliable source examples#arXiv preprints and conference abstracts. PrimeHunter (talk) 13:09, 12 March 2017 (UTC)

The correctness of Walter Wyss's proof is disputed by Ruslan Sharipov in arXiv:1704.00165. Chris Thompson (talk) 14:23, 22 May 2018 (UTC)

We've already gone a circle of this before; see the article history from March 2017 as well as e.g. this thread on my talkpage. It was briefly and regrettably added to the article then, and removed within two weeks; there is no reason at all to think the arXiv posting is a valid proof of anything. --JBL (talk) 14:33, 22 May 2018 (UTC)

https://arxiv.org/abs/2203.01149 Another unreviewed proof, by Natalia V. Aleshkevich in March 2022. Good thing I read this talk page, I was about to edit the article. 104.246.130.239 (talk) 17:30, 12 November 2022 (UTC)

Proposed merge of Cuboid conjectures into Euler brick

Too specific a subtopic to stand alone. –LaundryPizza03 (dc̄) 15:08, 2 October 2020 (UTC)

• Support. This would avoid to have a article based on two WP:primary sources by the same author, published in a minor jounal. Normally, this would be sufficient for deleting the article, but, as this is an interesting result, the merge is a good compromise. D.Lazard (talk) 15:51, 2 October 2020 (UTC)
• Support. Good solution to otherwise dubious notability.--agr (talk) 16:12, 2 October 2020 (UTC)
• Sure -- that's clearly not enough for its own article, but it's good enough for a couple of sentences here. --JBL (talk) 13:03, 6 October 2020 (UTC)

  Y Merger complete. Klbrain (talk) 11:01, 1 May 2021 (UTC)

Naive question

Article says "Only recently have cuboids in complex numbers become known." However, I have a query, what are Euler bricks called if they are expressed in https://en.wikipedia.org/wiki/Non-integer_base_of_numeration ?

I've tried googling, but, maybe I am just really naive, and there is no such thing?

Thank you to any Wikipedian who can answer this query. 49.185.41.142 (talk) 20:24, 2 May 2022 (UTC)

The property of a length to be an integer or not does not depend on the base of numeration that is used to represent numbers. So, "expressing Euler bricks in non-integer base of numeration" is nonsensical. D.Lazard (talk) 06:43, 3 May 2022 (UTC)