[sub-project] Perfect cuboid

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Zak
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Re: [sub-project] Perfect cuboid

#181 Ungelesener Beitrag von Zak » 24.01.2018 19:25

x3mEn
Because all members of the original expression have a second degree.
You like a sixth-degree polynomial - excellent, it must be brought to the symmetric form of the multiply of the same expression itself on itself: (a^2+b^2)(a^2+c^2)(b^2+c^2)(a^2+b^2+c^2)=(expression)*(expression) by using only members a^2, b^2, c^2 or by using them in any degree for your taste, if this will be easier for you. So we will get the expression which have natural value and then square root of multiply (expression)*(expression) will give the natural solution on the 100 percents. As a result, we should get clear (expression).

x3mEn
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Re: [sub-project] Perfect cuboid

#182 Ungelesener Beitrag von x3mEn » 24.01.2018 20:11

Ok, you don't know how to represent the product of four as a full square. And so what?
Is it your proof, "I don't know how to express, so it can never be!" ?

Yes, the product of four is a sixth-degree polynomial, you can easily make sure, if you open parentheses:
CodeCogsEqn.gif
CodeCogsEqn.gif (5.12 KiB) 5428 mal betrachtet
Trying to express sixth-degree polynomial as a qudratic polynomial you just reduce all to the solution of the Diophantine equation of the 6th degree.
But why the (expression)^2 cannot be, for example, of the form of
CodeCogsEqn3.gif
CodeCogsEqn3.gif (2.59 KiB) 5427 mal betrachtet
At least it looks more realistic because this is also a sixth-degree polynomial.

Zak
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Re: [sub-project] Perfect cuboid

#183 Ungelesener Beitrag von Zak » 24.01.2018 21:54

00.gif
00.gif (11.62 KiB) 5421 mal betrachtet

x3mEn
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Re: [sub-project] Perfect cuboid

#184 Ungelesener Beitrag von x3mEn » 24.01.2018 22:37

No proof, no conclusions. It's just mind games.

Zak
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Re: [sub-project] Perfect cuboid

#185 Ungelesener Beitrag von Zak » 24.01.2018 22:56

It's nice to play these games with a person who has a beautifull mind.

x3mEn
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Re: [sub-project] Perfect cuboid

#186 Ungelesener Beitrag von x3mEn » 25.01.2018 16:43

Zak,
I figured out how to show the absurdity of your "proof"

d^2 = a^2 + b^2
Now I'll "prove" that the face diagonal d can never be integer.
d = sqrt(a^+b^2), let's suppose d is integer, it's possible only when a^2+b^2 (1) is a full square:
a^2+b^2=(expression)(expression)
Let's expression = m*a + n*b
(m*a + n*b)^2 = m^2*a^2 + 2mnab + n^2*b^2 (2)
As far as a^2+b^2 has koef 1 before a^2 and b^2 and koef 0 before ab, matching (2) with (1) we will get a system of equations:
m^2 = 1
n^2 = 1
2mn = 0
As we can see, the system of equations has no solutions for m and n, it means a^2+b^2 cannot be expressed as (m*a + n*b)^2.
As far as a^2+b^2 cannot be expressed as a full square, so the face diagonal d can never be integer.


Try to find a logical trap in my "proof" and you'll understand how absurdly is yours.

Zak
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Re: [sub-project] Perfect cuboid

#187 Ungelesener Beitrag von Zak » 22.02.2018 14:48

x3mEn

Try:
Снимок.GIF
Снимок.GIF (12.46 KiB) 5091 mal betrachtet
Снимок2.GIF
Снимок2.GIF (13.29 KiB) 5088 mal betrachtet

Zak
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Re: [sub-project] Perfect cuboid

#188 Ungelesener Beitrag von Zak » 29.05.2018 17:49

Hi!
a^2+b^2=d^2
a^2+c^2=e^2
b^2+c^2=f^2
a^2+b^2+c^2=g^2

f^2=2a^2+(b^2+c^2-2a^2)
g^2=3a^2+(b^2+c^2-2a^2)

Let: (b^2+c^2-2a^2)=x, then:

gf=sqr((3a^2+x)(2a^2+x)), so under square root expression is not a square, therefore gf - not integer
g - not integer

x3mEn
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Re: [sub-project] Perfect cuboid

#189 Ungelesener Beitrag von x3mEn » 30.05.2018 09:48

Let:
a = 448
b = 495
c = 840

f = sqrt(495^2+840^2) = 975
g = sqrt(448^2+495^2+840^2) = 1073

x = b^2+c^2-2a^2 = 495^2+840^2-2*448^2 = 549217
sqrt((3a^2+x)(2a^2+x)) = sqrt((3*448^2+549217)(2*448^2+549217)) = sqrt(1151329*950625) = sqrt(1046175^2) = 1046175

gf = 975*1073 = 1046175

Zak
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Re: [sub-project] Perfect cuboid

#190 Ungelesener Beitrag von Zak » 30.05.2018 20:26

x3mEn
What about gdef?
Euler brick has integer: d,e,f

de=sqr((3a^2+(x-c^2))(3a^2+(x-b^2)))
gf=sqr((3a^2+x)(2a^2+x))

At least one of this two under square root expressions not a square

de - not integer or gf - not integer

gf/de - irrational
01.GIF
01.GIF (48.98 KiB) 3720 mal betrachtet

Zak
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Re: [sub-project] Perfect cuboid

#191 Ungelesener Beitrag von Zak » 16.07.2018 20:32

Hi!
May be following information helps you...
In the figure, yellow color conditionally shows half of the sweep of the Euler brick abc. We show that the length of the space diagonal in the Euler brick is not an integer. So. Simply look at the last equation on the figure:
fig01.jpg
fig01.jpg (97.99 KiB) 2961 mal betrachtet
www.wolframalpha.com_the equation solutions

You may also find an y coordinate which possible for a point A, and then find the lenght CA by dint of coordinates.
Hence the coordinate x of the point A, which is the end of the distance CA has no solutions in integers for the Euler cuboid. Accordingly, after substitution, the coordinate y also/or has no integer solutions, that is: x, y∉N. Since the point C (-a; -b), which is the end of the same distance CA has integer coordinates, then the length CA, and, consequently, the length of the equal space diagonal of a given Euler brick can not be expressed as an integer.

x3mEn
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Re: [sub-project] Perfect cuboid

#192 Ungelesener Beitrag von x3mEn » 18.07.2018 21:49

CodeCogs.gif
CodeCogs.gif (8.62 KiB) 2928 mal betrachtet
Why do you never check your galactic hypotheses on practice?

I guess you know the story about Landau and "farmatists":

Landau was very much bothered by "farmatists" — dilettantes trying to prove Fermat's Great Theorem and get a major award for this proof. In order not to be distracted from the main work, Landau ordered several hundred forms with the following text:

"Dear ...! Thank you for the manuscript sent by you with the proof of Fermat's Great Theorem. The first error is on page ... in the line ... "

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