[sub-project] Perfect cuboid

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Re: [sub-project] Perfect cuboid

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).
Zak
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Re: [sub-project] Perfect cuboid

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 (5.12 KiB) Viewed 3751 times

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 (2.59 KiB) Viewed 3750 times

At least it looks more realistic because this is also a sixth-degree polynomial.
x3mEn
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Re: [sub-project] Perfect cuboid

00.gif (11.62 KiB) Viewed 3744 times
Zak
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Re: [sub-project] Perfect cuboid

No proof, no conclusions. It's just mind games.
x3mEn
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Re: [sub-project] Perfect cuboid

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

Posts: 106
Joined: 04.11.2017 12:27

Re: [sub-project] Perfect cuboid

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.
x3mEn
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Re: [sub-project] Perfect cuboid

x3mEn

Try:

Снимок.GIF (12.46 KiB) Viewed 3414 times

Снимок2.GIF (13.29 KiB) Viewed 3411 times
Zak
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Posts: 106
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Re: [sub-project] Perfect cuboid

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
Zak
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Joined: 04.11.2017 12:27

Re: [sub-project] Perfect cuboid

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
x3mEn
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Re: [sub-project] Perfect cuboid

x3mEn
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 (48.98 KiB) Viewed 2043 times
Zak
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Re: [sub-project] Perfect cuboid

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 (97.99 KiB) Viewed 1284 times

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.
Zak
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Posts: 106
Joined: 04.11.2017 12:27

Re: [sub-project] Perfect cuboid

CodeCogs.gif (8.62 KiB) Viewed 1251 times

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 ... "
x3mEn
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