[sub-project] Perfect cuboid

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Re: [sub-project] Perfect cuboid

Unread postby 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).
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Re: [sub-project] Perfect cuboid

Unread postby 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) Viewed 2081 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
CodeCogsEqn3.gif (2.59 KiB) Viewed 2080 times

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

Unread postby Zak » 24.01.2018 21:54

00.gif
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Re: [sub-project] Perfect cuboid

Unread postby x3mEn » 24.01.2018 22:37

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

Unread postby Zak » 24.01.2018 22:56

It's nice to play these games with a person who has a beautifull mind.
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Re: [sub-project] Perfect cuboid

Unread postby 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.
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Re: [sub-project] Perfect cuboid

Unread postby Zak » 22.02.2018 14:48

x3mEn

Try:

Снимок.GIF
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Снимок2.GIF
Снимок2.GIF (13.29 KiB) Viewed 1741 times
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Re: [sub-project] Perfect cuboid

Unread postby 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
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Re: [sub-project] Perfect cuboid

Unread postby 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
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Re: [sub-project] Perfect cuboid

Unread postby 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
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