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Is there any consistent geometry in which $a^3+b^3=c^3$ holds for right-angled triangles with 'hypothenuse' $c$?

Is there any natural generalization of pythagoras theorem to higher dimensions?

Is there any geometric shape with $3$ sides, with areas $a,b,c$ and $a^3+b^3=c^3$

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I presume you mean a 3-dimensional geometric shape with polygons for faces. Such a thing needs at minimum 4 faces (simplex). As for your first question, take $\mathbb{R}^2$ with the $\ell_3$ norm. If the previous sentence doesn't make sense to you, then MO is not the site you are looking for as it is for research-level mathematicians. math.stackexchange.com welcomes mathematical questions at all levels. –  David Roberts Oct 10 '11 at 4:40
    
Also posted to m.se, math.stackexchange.com/questions/71271/… –  Gerry Myerson Oct 10 '11 at 4:45
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closed as off topic by David Roberts, Gerry Myerson, Andres Caicedo, HJRW, David Loeffler Oct 10 '11 at 7:15

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1 Answer

Is there any natural generalization of pythagoras theorem to higher dimensions?

From http://www.cs.bc.edu/~alvarez/NDPyt.pdf :

... we will restrict our attention to the analogs of right triangles. We will call these orthogonal tetrahedrons. An orthogonal tetrahedron is any tetrahedron which has a vertex at which three faces meet at right angles to each other ... we refer to [these faces] as the orthogonal faces, and we refer to the remaining face as the opposing face ...

Let $A$ be the area of the opposing face, and let $A_1 , A_2 , A_3$ be the areas of the orthogonal faces of a given orthogonal tetrahedron. Then the following relation holds: $A^2 = A_1^2 + A_2^2 + A_3^2$

Next we address the n-dimensional case ...

Let $A$ be the area of the opposing face, and let $A_1, A_2, ..., A_n$ be the areas of the orthogonal faces of a given orthogonal n-simplex. Then the following relation holds:

$$A^2 = \sum_{j=1}^n A_j^2$$

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This answer relies on readers downloading and reading this pdf you link to. Good answers are self-contained. Would you care to summarise the important points? –  David Roberts Oct 10 '11 at 4:46
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