# Graphically display $a^3+b^3+c^3 = d^3$

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.
-
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"? – joriki Dec 22 '11 at 15:11
Thanks. I had a hard time formulating the question. – Hidde Dec 22 '11 at 22:04

It's a nice idea. I don't know of any way to make it work with the cubes, but you might like this interpretation of the triangle whose area you calculated. That area is (with $\vec a,\vec b,\vec c$ the vectors corresponding to the three sides)

$$\frac12\left|\left(\vec a-\vec b\right)\times\left(\vec b - \vec c\right)\right|\;,$$

so its square comes out as

$$\left(\frac12\vec a\times \vec b\right)^2+\left(\frac12\vec b\times \vec c\right)^2+\left(\frac12\vec c\times \vec a\right)^2\;,$$

which is the sum of the squares of the areas of the three other triangles formed by the three sides.

(I found this idea here, but the proofs there are unnecessarily complicated.)

This generalizes to higher dimensions, with the cross product replaced by the exterior product.

-