[sub-project] Perfect cuboid

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Re: [sub-project] Perfect cuboid

Unread postby Zak » 25.10.2018 21:00

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?
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Re: [sub-project] Perfect cuboid

Unread postby Zak » 26.10.2018 06:23

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.
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Re: [sub-project] Perfect cuboid

Unread postby Zak » 26.10.2018 19:32

x3mEn
Is it fair?
m5.JPG
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Or may be two perfect bricks exists? How to explain it?
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Re: [sub-project] Perfect cuboid

Unread postby Zak » 27.10.2018 13:01

I think it is possible only in case D=0, d1=d2:
m6.JPG
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Re: [sub-project] Perfect cuboid

Unread postby x3mEn » 27.10.2018 16:42

Zak,
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Re: [sub-project] Perfect cuboid

Unread postby Zak » 27.10.2018 18:30

x3mEn
Try:
m7.JPG
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Re: [sub-project] Perfect cuboid

Unread postby Zak » 28.10.2018 14:53

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
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Re: [sub-project] Perfect cuboid

Unread postby x3mEn » 29.10.2018 22:33

Zak,
CodeCogsSEFX.gif
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Re: [sub-project] Perfect cuboid

Unread postby Zak » 30.10.2018 19:08

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.
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Re: [sub-project] Perfect cuboid

Unread postby x3mEn » 30.10.2018 20:52

Zak,
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Re: [sub-project] Perfect cuboid

Unread postby Zak » 04.12.2018 21:12

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.
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Re: [sub-project] Perfect cuboid

Unread postby Zak » 07.12.2018 22:44

x3mEn
Is it clear:
e1.JPG
e1.JPG (48.58 KiB) Viewed 176 times

e2.JPG
e2.JPG (80.56 KiB) Viewed 176 times

?
Do you agree?
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