## [sub-project] Perfect cuboid

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Zak
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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 hat geschrieben: 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) 5154 mal betrachtet
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) 5134 mal betrachtet

x3mEn
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### Re: [sub-project] Perfect cuboid

Zak,
CodeCogsDEFX.gif (10.55 KiB) 5120 mal betrachtet

Zak
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### Re: [sub-project] Perfect cuboid

x3mEn
Try:
m7.JPG (12.17 KiB) 5119 mal betrachtet

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) 5070 mal betrachtet

x3mEn
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### Re: [sub-project] Perfect cuboid

Zak,
CodeCogsSEFX.gif (1.28 KiB) 5039 mal betrachtet

Zak
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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.

x3mEn
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### Re: [sub-project] Perfect cuboid

Zak,
CodeCogsSEFX2.gif (2.42 KiB) 4992 mal betrachtet

Zak
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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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### Re: [sub-project] Perfect cuboid

x3mEn
Is it clear:
e1.JPG (48.58 KiB) 4183 mal betrachtet
e2.JPG (80.56 KiB) 4183 mal betrachtet
?
Do you agree?