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

Verfasst:

**13.09.2018 01:52**x3mEn

Hi!

Hi!

die erste Adresse für Distributed Computing

https://www.rechenkraft.net/forum/

https://www.rechenkraft.net/forum/viewtopic.php?f=56&t=11559

Seite **18** von **22**

Verfasst: **13.09.2018 01:52**

x3mEn

Hi!

Hi!

Verfasst: **13.09.2018 20:58**

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.

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.

Verfasst: **14.09.2018 07:36**

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!

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!

Verfasst: **15.09.2018 14:16**

You muss "could" and "should".Zak hat geschrieben:x3mEn

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

Yes, theoretically X could be integer, but does it should be? No.

At least until you (or someone else) prove that it shoud.

Verfasst: **15.09.2018 14:51**

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

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

Verfasst: **22.09.2018 23:19**

x3mEn

Verfasst: **24.09.2018 20:59**

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.

Verfasst: **24.09.2018 23:24**

x3mEn

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

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

Verfasst: **25.09.2018 07:26**

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

(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

Verfasst: **01.10.2018 20:52**

You've got a mistake.x3mEn hat geschrieben: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

For your case all be the same:

As I said, the proof is true for the invariant.

Are you still searching?

Verfasst: **02.10.2018 10:11**

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.

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.

Verfasst: **02.10.2018 16:36**

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