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x3mEn
if g - odd:
mi must be odd
ni must be even
mi>ni
ki must be odd
Try to use these conditions. You can't to use n3=1. You used the fact that the unit in any degree is equal to one.

x3mEn
First. All actions for the all eight cases will be the same because they are commutative with accurancy to designations.
Second.

x3mEn hat geschrieben: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.

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

You've got a mistake.
For your case all be the same:
As I said, the proof is true for the invariant.
Are you still searching?

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

x3mEn

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...

x3mEn
Hi!

x3mEn
How do you thing about this:

O.K. Thanks. So if we have an irrational space diagonal then at least one of coordinate will be irrational. I will show you the invariance test. I have write up a square to you: (a^2+b^2)+c^2 (c^2+a^2)+b^2 (b^2+c^2)+a^2 It's a square. You can see the vertical rows and horizontal lines in it. And eve...

If I am not mistaken

Excellent, but let's move to the triple dimension. Now do the same with the other two face diagonals for Euler's brick. Then imagine that we have placed the three halves of the sweep so that the points C and A coincide for all three cases, since the length of the space diagonal is the same. As a res...