[sub-project] Perfect cuboid

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Re: [sub-project] Perfect cuboid

Even if x=sqr(16a^2b^2c^2(a^2+b^2+c^2))?
Try to get integer x=4abc*sqr(a^2+b^2+c^2) by using real(including irrational, of course) values.
I can't. May be you can do it?
Zak
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Re: [sub-project] Perfect cuboid

x3mEn wrote:If you will find integer d,e,f,x and calculate a,b,c,g based on d,e,f, they (I mean a,b,c,g) will not necessarily be integer, some of them or even entire all can turn to irrational. Despite the fact that x^2=16a^2*b^2*c^2*g^2 will be TRUE.
Because a product of irrational numbers could be an integer, and not just an integer, but a full square. Easy.

I don't think so. I think it generally impossible to find integers d1,d2,e,f,x. Never.
Zak
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Re: [sub-project] Perfect cuboid

x3mEn
Is it fair?
m5.JPG (32.23 KiB) Viewed 1077 times

Or may be two perfect bricks exists? How to explain it?
Zak
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Re: [sub-project] Perfect cuboid

I think it is possible only in case D=0, d1=d2:
m6.JPG (20.67 KiB) Viewed 1057 times
Zak
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Re: [sub-project] Perfect cuboid

Zak,
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x3mEn
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Re: [sub-project] Perfect cuboid

x3mEn
Try:
m7.JPG (12.17 KiB) Viewed 1042 times
Zak
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Re: [sub-project] Perfect cuboid

x3mEn
The discriminant cannot be zero, I made a mistake above, there will always be two roots.
Maybe you will be interested to solve the last expression in integers, it will be the geometric mean, then both roots will be integers, if I don't mistake:
g.JPG (18.64 KiB) Viewed 993 times
Zak
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Re: [sub-project] Perfect cuboid

Zak,
CodeCogsSEFX.gif (1.28 KiB) Viewed 962 times
x3mEn
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Re: [sub-project] Perfect cuboid

x3mEn
Are you looking for complex values of the face diagonal? It is obvious that if integer values are needed, not complex values, when x^2<4e^4f^4. Insert this condition and try the same one more time, please.
Zak
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Re: [sub-project] Perfect cuboid

Zak,
CodeCogsSEFX2.gif (2.42 KiB) Viewed 915 times
x3mEn
XBOX360-Installer

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Re: [sub-project] Perfect cuboid

x3mEn
Hi! Let me invite you to this discussion:
http://mathhelpplanet.com/viewtopic.php?f=57&t=62956
I think there is not roots among the divisors of the free member of two last equations.
Zak
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Posts: 105
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Re: [sub-project] Perfect cuboid

x3mEn
Is it clear:
e1.JPG (48.58 KiB) Viewed 106 times

e2.JPG (80.56 KiB) Viewed 106 times

?
Do you agree?
Zak
Prozessor-Polier

Posts: 105
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