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As a kind of 'addition' to Fermats Last Theorem, a friend of mine has come up with a different idea. Let $$a^3+b^3+c^3 = d^3$$ with $(a,b,c,d): a,b,c,d \in \mathbb{N}$.

We were discussing Pythagoras: how you could constuct a triangle from two sides ($a$, $b$) of two squares ($a^2$, $b^2$) (at a right angle), which results in the third side of the triangle, which is also the side of the third square ($c^2$).

How could we do that with a dimension higher, if we say that $a$, $b$ and $c$ are sides of cubes, and we position the cubes in a 3D (x, y, z) coordinate system. How will the fourth cube (with side $d$) come from the three cubes?

We know that the set $a = 3$, $b = 4$, $c = 5$ and $d = 6$ work out, and we've tried putting all of the sides on an axis each, and calculating the area of the triangle the three points make. This made sense to us: two sides of two squares make a line, so three sides of three cubes, should make an area. The area of the triangle was a dead end (as far as we could find, the area ($\frac{\sqrt{769}}{2}$)has nothing to do with the number 6), and we couldn't think of anything more.

The questions:

  • Is there any way to position the three cubes, that they make sense in the same way the squares in a Pythagoras triangle make sense?
  • How does the fourth cube come from the three cubes, just like in a Pythagoras triangle.
  • 0
    I think you mean "two sides of a squares" -> "two sides of two squares" and "How does the third cube" -> "How does the fourth cube"?2011-12-22
  • 0
    Thanks. I had a hard time formulating the question.2011-12-22

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