## [sub-project] Perfect cuboid

Fehler und Wünsche zum Projekt yoyo@home
Bugs and wishes for the project yoyo@home

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

x3mEn
Hi!

000.JPG (72.17 KiB) Viewed 1606 times
Zak
Prozessor-Polier

Posts: 107
Joined: 04.11.2017 12:27

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

Try
b = 1443
c = 1800
d = 1595
e = 1924
f = 2307
g = 2405
and check (5) and (6):

2405 = sqrt(1595^2 + 1800^2) — true
1924 = sqrt(1595^2 + (1800^2 - 1443^2)) — true

x^2 = c^2 - b^2 = 1800^2 - 1443^2 = 1157751
x = sqrt(1157751) = 3*sqrt(128639) ∉ Z
and I don't see any reason why x should be necessarily integer, there are infinitely many examples when (b, c, d, e, f, g) are integers, and (5), (6) and even b^2+c^2 = f^2 (7) expressions are true too.
x3mEn
XBOX360-Installer

Posts: 99
Joined: 20.03.2011 22:23

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

x3mEn
I know about this cases, when sum of square of integer and integer is a square of integer, but you didn't anderstand to me.
The expression c^2+b^2i^2=x^2=c^2-b^2, where i^2=-1 is a triple. When we introduced the integer coefficient (-1) into the integer values expression c^2+b^2=f^2, there was no reason to make X non-integer. What made X become non-integer? Nothing. And then we'll see complex value and the integer value of the face diagonal the same time: e=sqrt(d^2+(c^2+b^2i^2)). And now also there this no reason to change integer values over non-integers: x=sqrt(f^2-2b^2)=sqrt(c^2+b^2i^2), but it's impossible!
Zak
Prozessor-Polier

Posts: 107
Joined: 04.11.2017 12:27

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

Zak wrote:x3mEn
there was no reason to make X non-integer. What made X become non-integer? Nothing.

You muss "could" and "should".
Yes, theoretically X could be integer, but does it should be? No.
At least until you (or someone else) prove that it shoud.
x3mEn
XBOX360-Installer

Posts: 99
Joined: 20.03.2011 22:23

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

It seems like you try to prove
if c^2+b^2=f^2 for integer c, b and f, than c^2-b^2 must be a full square of integer X, because "there was no reason to make X non-integer".
This is a complete nonsense.
X could be integer or real, depends on.
The question is equal to solution of the system of Diophantine equations
a^2 + b^2 = c^2
a^2 - b^2 = d^2
x3mEn
XBOX360-Installer

Posts: 99
Joined: 20.03.2011 22:23

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

x3mEn
00.JPG (65.53 KiB) Viewed 1493 times
Zak
Prozessor-Polier

Posts: 107
Joined: 04.11.2017 12:27

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

When the Pythagorean triple has m^2-n^2 > 2mn then a = 2mn, b = m^2-n^2 and all your "strong" proof falls down.
x3mEn
XBOX360-Installer

Posts: 99
Joined: 20.03.2011 22:23

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

x3mEn
The variables a, b, c are interchangeable and the proof is valid in the invariant. You are not right. Are you still searching?
Zak
Prozessor-Polier

Posts: 107
Joined: 04.11.2017 12:27

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

Lets
(a, b, c) = (44, 117, 240)
(d, e, f) = (125, 244, 267)

a^2 + b^2 = d^2
a^2 + c^2 = e^2
b^2 + c^2 = f^2

44^2 + 117^2 = 125^2
44^2 + 240^2 = 244^2
117^2 + 240^2 = 267^2

(44, 117, 125) is a primitive Pythagorian triple and can be expressed by (m, n) as
m = 11, n = 2
m^2 - n^2 = 11^2 - 2^2 = 117 = b
2*m*n = 2*13*4 = 44 = a
m^2 + n^2 = 11^2 + 2^2 = 125 = d

As far as you express 2g^2 as a sum of d^2 + e^2 + b^2 + c^2 = 125^2 + 244^2 + 117^2 + 240^2
, you cannot arbitrarily change b to a (117 to 44) just because a=2mn.
When you swap values of a and b you also have to swap values of e and f:
2g^2 = d^2 + f^2 + a^2 + c^2 = 125^2 + 267^2 + 44^2 + 240^2

(a, b, c) = (117, 44, 240)
(d, e, f) = (125, 267, 244)

117^2 + 44^2 = 125^2 = d^2
117^2 + 240^2 = 267^2 = e^2
44^2 + 240^2 = 240^2 = f^2

2g^2 = d^2 + e^2 + b^2 + c^2 = 125^2 + 267^2 + 44^2 + 240^2
2g^2 = d^2 + f^2 + a^2 + c^2 = 125^2 + 244^2 + 117^2 + 240^2
x3mEn
XBOX360-Installer

Posts: 99
Joined: 20.03.2011 22:23

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

x3mEn wrote:Lets
(44, 117, 125) is a primitive Pythagorian triple and can be expressed by (m, n) as
m = 11, n = 2
m^2 - n^2 = 11^2 - 2^2 = 117 = b
2*m*n = 2*13*4 = 44 = a

You've got a mistake.
For your case all be the same:

00.JPG (66.25 KiB) Viewed 1330 times

As I said, the proof is true for the invariant.
Are you still searching?
Zak
Prozessor-Polier

Posts: 107
Joined: 04.11.2017 12:27

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

Yes, I've got a mistake.
m = 11, n = 2
2*m*n = 2*11*2 = 44 = a
So, it nothing changes.
All other sentences remain the same: you cannot arbitrarily swap values of b to a without swap values of e and f.
x3mEn
XBOX360-Installer

Posts: 99
Joined: 20.03.2011 22:23

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

x3mEn wrote:you cannot arbitrarily swap values of b to a without swap values of e and f.

I see no obstacles. I don't understand what you mean. I showed to you analogically proof for case when a=2mn b=m^2-n^2 either a=m^2-n^2 b=2mn.
Zak
Prozessor-Polier

Posts: 107
Joined: 04.11.2017 12:27

PreviousNext