Seite 21 von 22

Re: [sub-project] Perfect cuboid

Verfasst: 25.10.2018 21:00
von Zak
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?

Re: [sub-project] Perfect cuboid

Verfasst: 26.10.2018 06:23
von Zak
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.

Re: [sub-project] Perfect cuboid

Verfasst: 26.10.2018 19:32
von Zak
x3mEn
Is it fair?
m5.JPG
m5.JPG (32.23 KiB) 2493 mal betrachtet
Or may be two perfect bricks exists? How to explain it?

Re: [sub-project] Perfect cuboid

Verfasst: 27.10.2018 13:01
von Zak
I think it is possible only in case D=0, d1=d2:
m6.JPG
m6.JPG (20.67 KiB) 2473 mal betrachtet

Re: [sub-project] Perfect cuboid

Verfasst: 27.10.2018 16:42
von x3mEn
Zak,
CodeCogsDEFX.gif
CodeCogsDEFX.gif (10.55 KiB) 2459 mal betrachtet

Re: [sub-project] Perfect cuboid

Verfasst: 27.10.2018 18:30
von Zak
x3mEn
Try:
m7.JPG
m7.JPG (12.17 KiB) 2458 mal betrachtet

Re: [sub-project] Perfect cuboid

Verfasst: 28.10.2018 14:53
von Zak
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
g.JPG (18.64 KiB) 2409 mal betrachtet

Re: [sub-project] Perfect cuboid

Verfasst: 29.10.2018 22:33
von x3mEn
Zak,
CodeCogsSEFX.gif
CodeCogsSEFX.gif (1.28 KiB) 2378 mal betrachtet

Re: [sub-project] Perfect cuboid

Verfasst: 30.10.2018 19:08
von Zak
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.

Re: [sub-project] Perfect cuboid

Verfasst: 30.10.2018 20:52
von x3mEn
Zak,
CodeCogsSEFX2.gif
CodeCogsSEFX2.gif (2.42 KiB) 2331 mal betrachtet

Re: [sub-project] Perfect cuboid

Verfasst: 04.12.2018 21:12
von Zak
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.

Re: [sub-project] Perfect cuboid

Verfasst: 07.12.2018 22:44
von Zak
x3mEn
Is it clear:
e1.JPG
e1.JPG (48.58 KiB) 1522 mal betrachtet
e2.JPG
e2.JPG (80.56 KiB) 1522 mal betrachtet
?
Do you agree?