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

x3mEn
Try:

If I was not mistaken, then it's followed:

x3mEn
I will help you next: Sorry, mistake was corrected.
Now everything is checked and true.

x3mEn Try to simplify the expression: sqr(a^2+b^2)+sqr(a^2+c^2)+sqr(b^2+c^2) to the desired form at the right part: q*sqr(a^{2}+b^{2}+c^{2}) Maybe you even get also to simplify sqr(u)+sqr(v)+sqr(w) which united to the single common radical Try to express the variable n only through variables u,v,w ,...

So I mean it's impossible to reduce left part with a sum of three radicals to right part which consist of single radical of the finaly expression:

Hi! How is your searching? )))

Hi! I found a proof: The cube abc lies in a rectangular coordinate system so that one of its vertices is the origin, and its edges lie on the coordinate axes in positive directions, respectively. So: g^2=a^2+b^2+c^2 is the equation of a sphere of radius g with the center at the origin. Let a,b,c,g a...

So

To be honest, very unlikely a perfect cuboid exists at all... g^2=d^2+e^2+f^2-a^2-b^2-c^2=(a^2+b^2-c^2)+(a^2+c^2-b^2)+(b^2+c^2-a^2)=a^2+b^2+c^2 So you will snatch the jackpot when you will find: If a<b<c then (a^2+b^2-c^2)<(a^2+c^2-b^2)<(b^2+c^2-a^2) a^2=(a^2+b^2-c^2) b^2=(a^2+c^2-b^2) c^2=(b^2+c^2...

We also can to consider the triangle d,e,f with acute angles and we can find cosinuses of them analogically... And finally we have three Pyphagorean triples, which may be described by two integer values of m and n , and if we inevitably have at least one edge even, thus all the parameters: a,b,c,d,e...

O.K.
x, f - are integer
g=sqr(f^2+x^2-2*f*x*cos(α)) where angle (α) - obtuse angle and angle (α) can't equal 120 degrees for all cases with each of three face diagonals, thus expression 2*f*x*cos(α) haven't integger value at least once of three cases and the space diagonal g will not integger.