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.