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Re: [sub-project] Perfect cuboid

Verfasst: 24.01.2018 19:25
von Zak
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).

Re: [sub-project] Perfect cuboid

Verfasst: 24.01.2018 20:11
von x3mEn
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) 8736 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) 8735 mal betrachtet
At least it looks more realistic because this is also a sixth-degree polynomial.

Re: [sub-project] Perfect cuboid

Verfasst: 24.01.2018 21:54
von Zak
00.gif
00.gif (11.62 KiB) 8729 mal betrachtet

Re: [sub-project] Perfect cuboid

Verfasst: 24.01.2018 22:37
von x3mEn
No proof, no conclusions. It's just mind games.

Re: [sub-project] Perfect cuboid

Verfasst: 24.01.2018 22:56
von Zak
It's nice to play these games with a person who has a beautifull mind.

Re: [sub-project] Perfect cuboid

Verfasst: 25.01.2018 16:43
von x3mEn
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.

Re: [sub-project] Perfect cuboid

Verfasst: 22.02.2018 14:48
von Zak
x3mEn

Try:
Снимок.GIF
Снимок.GIF (12.46 KiB) 8399 mal betrachtet
Снимок2.GIF
Снимок2.GIF (13.29 KiB) 8396 mal betrachtet

Re: [sub-project] Perfect cuboid

Verfasst: 29.05.2018 17:49
von Zak
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

Re: [sub-project] Perfect cuboid

Verfasst: 30.05.2018 09:48
von x3mEn
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

Re: [sub-project] Perfect cuboid

Verfasst: 30.05.2018 20:26
von Zak
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) 7028 mal betrachtet

Re: [sub-project] Perfect cuboid

Verfasst: 16.07.2018 20:32
von Zak
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) 6269 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.

Re: [sub-project] Perfect cuboid

Verfasst: 18.07.2018 21:49
von x3mEn
CodeCogs.gif
CodeCogs.gif (8.62 KiB) 6236 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 ... "