[sub-project] Perfect cuboid

[Zak]

x3mEn
Try:

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01.gif

[Zak]

So, at this new formula (above) of spase diagonal for any Euler brick we have:

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Is it clear?

[Zak]

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[Zak]

x3mEn
Do you agree?

[x3mEn]

Zak,
the only thing you achieved, you derived that for given natural d, e, f and g the next equation must be true:

CodeCogsEqn.gif

Let's check for our already well-known Edge cuboid:

CodeCogsEqn3.gif

[x3mEn]

CodeCogsEqn4.gif

So your conclusion that for given natural d, e, f and g there is no natural solution for m and n — completely wrong.

[Zak]

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[lloyd]

...

[Zak]

x3mEn

00.gif

01.gif

[x3mEn]

CodeCogsEqn9.gif

[Zak]

x3Men
One more time: "we assume that in Euler brick..."
It's very well that you can find numbers, but the edge c is not integer in you example. Need not to search solution, need only simplify algebraic expression by combination law (assotiation law) to the desired form.

For example on manner: 4a^4+16b^4+16a^2b^2=(2a^2+4b^2)(2a^2+4b^2)

[x3mEn]

Your "proof" contains only d, e, f, g and m, n, k naturalness requirement and nothing about a, b, c
I brought a counter-example where d, e, f, g are natural and natural m, n, k also exist, so your conclusion that d*e*f*g can never be natural is wrong.
You hadn't explain in a proper way why a product of (a^2+b^2)(a^2+c^2)(b^2+c^2)(a^2+b^2+c^2) can never be a full square.
And also I didn't catch your point, why the upper product has to be represented exactly as a quadratic polynomial (ua^2+vb^2+wc^2)^2.
The upper product is a polynomial of the sixth degree. Why did you decide to simplify it to quadratic polynomial — this is an enigma for me.