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The contents of the Cuboid conjectures page were
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Euler brick on 1 May 2021. For the contribution history and old versions of the redirected page, please see its history; for the discussion at that location, see
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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)reply
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)reply
Did you study the proof? Or do you base your assumption on the fact "It has not been published on journals"?
Actually it does not make a difference whether or not I read the proof, as that would be considered original research. The question I have is was it reviewed by others. My understanding is that it might not have been since research gate is not known for reviewing the works published there. If the proof were published in a mathematics journal, then it would have been reviewed before publishing.
TheRingess (
talk)
18:51, 18 January 2023 (UTC)reply
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)reply
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)reply
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)reply
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)reply
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."
If you are referring to the above discussion about primitive vs non-primitive Euler Bricks, your equation should be something like where . Nothing to prove there.
TheRingess (
talk)
04:09, 20 March 2013 (UTC)reply
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)reply
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)reply
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)reply
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)reply
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)reply
I see. The paper shows this method of constructing a perfect parallelepiped from Pythagorean triple cannot yield perfect cuboid, and even if it showed this applies to all Pythagorean triples, Pythagorean triples don't cover the whole space of perfect parallelepipeds, so the claim of proof of nonexistence of perfect cuboid is wrong by incompleteness.
104.246.130.239 (
talk)
05:59, 5 December 2022 (UTC)reply
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)reply
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
/info/en/?search=Non-integer_base_of_numeration ?
I've tried googling, but, maybe I am just really naive, and there is no such thing?
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)reply